Matter Waves: de Broglie Wavelength

Two particles A and B of masses $m_1$ and $m_2$ ($m_1 > m_2$) have the same de-Broglie wavelength, then

The de-Broglie wavelength $\lambda$ of a particle

The kinetic energy of a free electron increases to 3 times the previous K.E. The ratio of new de- Broglie wavelength to previous de-Broglie wavelength is

The ratio of de-Broglie wavelength of an $\alpha$-particle and proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$. The value of $m$ is

The proton and $\alpha$ - particle are accelerated through same potential difference. Then the ratio of the de-Broglie wavelength of proton and $\alpha$ - particle is (mass of $\alpha$-particle is $4$ times mass of proton, charge of $\alpha$-particle is $2$ times charge of proton)

If the kinetic energy of a particle is increased to 16 times, the percentage change in the de Broglie wavelength of a particle is

An electron is travelling with a velocity $v$ in free space and when it enters a medium, its velocity is reduced by $20\%$. The de Broglie wavelength of electron in the medium is $\alpha\lambda_0$, where $\lambda_0$ is its de Broglie wavelength in free space. The value of $\alpha$ is ______.

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases to $2\lambda$. The value of $(V_1/V_2)$ is equal to

The de Broglie wavelength for an electron accelerated through the potential difference of $V_1$ volt is $\lambda_1$. When the potential difference is changed to $V_2$ volt, the associated de Broglie wavelength is increased by $50\%$. If $(V_1/V_2) = (9/\alpha)$, then the value of $\alpha$ is ______。

A particle having electric charge $3 \times 10^{-19}\text{ C}$ and mass $6 \times 10^{-27}\text{ kg}$ is accelerated by applying an electric potential of 1.21 V.

Wavelength of the matter wave associated with the particle is $\alpha \times 10^{-12}\text{ m}$. The value of $\alpha$ is ______. (Take Planck's constant $= 6.6 \times 10^{-34}\text{ J s}$)

The ratio of the accelerating potentials required to accelerate (i) an $\alpha$-particle and (ii) a proton to have the same de Broglie wavelength associated with them is

(mass of $\alpha$ - particle = $6.4 \times 10^{-27}$ kg, mass of proton = $1.6 \times 10^{-27}$ kg)

The de Broglie wavelength associated with an electron accelerated through a potential difference V is $\lambda_\mathrm{e}$ and the de Broglie wavelength associated with a proton accelerated through the same potential difference is $\lambda_\mathrm{p}$. If their corresponding masses are $m_\mathrm{e}$ and $m_\mathrm{p}$, respectively, then the ratio of their de Broglie wavelengths $\left(\frac{\lambda_e}{\lambda_p}\right)$ is ______.

The ratio of de Broglie wavelength of a deuteron with kinetic energy $E$ to that of an alpha particle with kinetic energy $2E$, is $n : 1$. The value of $n$ is ______.

(Assume mass of proton $=$ mass of neutron)

The ratio of de-Broglie wavelengths of molecules of hydrogen and helium which are at temperatures $27^\circ C$ and $127^\circ C$ respectively is ($m_{He} = 2m_{H_2}$)

The de Broglie wavelength of an oxygen molecule at $27^\circ\text{C}$ is $x \times 10^{-12}\text{ m}$. The value of $x$ is (take Planck's constant $= 6.63 \times 10^{-34}\text{ J s}$, Boltzmann constant $= 1.38 \times 10^{-23}\text{ J/K}$, mass of oxygen molecule $= 5.31 \times 10^{-26}\text{ kg}$)

The wavelength '$\lambda$' of photon and the de-Broglie wavelength of an electron have same value. The ratio of kinetic energy of electron to the energy of a photon is (m=mass of electron, h = Planck's constant, c = velocity of light)

The energy of a photon is equal to the kinetic energy of a proton. If $\lambda_1$ is the de-Broglie wavelength of a proton, $\lambda_2$ is the wavelength associated with the photon and if the energy of the photon is $E$ then $(\lambda_1/\lambda_2)$ is proportional to

An electromagnetic wave of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have de-Broglie wavelength $\lambda_1$, then

According to de-Broglie hypothesis, the ratio of the wave-length of a photon and that of an electron having same energy 'E' is (m = mass of electron, c = velocity of light)

A photon and an electron have equal energy E. The ratio of $\lambda$(electron) to $\lambda$(photon) is proportional to

A photon and an electron, each of 20 eV energy, move in free space. The ratio of linear momentum of electron $p_e$ to that of photon $p_{Ph}, \frac{p_e}{p_{Ph}}$ is:

[Take speed of light $= 3 \times 10^{8}\text{ ms}^{-1}$, charge of electron $= -1.6 \times 10^{-19}\text{ C}$ and mass of electron $= 9 \times 10^{-31}\text{ kg}$ ]

An electron is moving with initial velocity $\vec{v} = a\hat{k}$ in a magnetic field $\vec{B} = b\hat{j}$. Its de-Broglie wavelength will

An electron is moving with initial velocity $\vec{v} = V_0\hat{j}$ in a magnetic field $\vec{B} = B_0\hat{i}$. Its de-Broglie wavelength will

A photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_0$. Expression for de-Broglie wavelength is (m = mass of electron, h = Planck's constant, C = Speed of light)

Light of wavelength '$\lambda$' which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with some velocity then stopping potential

An electron of mass $m$ and charge $e$ initially at rest gets accelerated by a constant electric field $E$. The rate of change of de-Broglie wavelength of the electron at time $t$ is

($h$ = Planck's constant)

An electron of mass $m$ is moving in an electric field $\vec{E} = -2E_o\hat{i}\ (E_o =$ constant $> 0)$, with an initial velocity $\vec{V} = v_o\hat{i}\ (v_o =$ constant $> 0)$. If $\lambda_o = \frac{h}{4mv_o}$, its de Broglie wavelength at time $t$ is ______.

( $e =$ charge of electron)

The de-Broglie wavelength $(\lambda)$ of a particle

If $\lambda$ and $K$ are de Broglie wavelength and kinetic energy, respectively, of a particle with constant mass. The correct graphical representation for the particle will be:

Electron beam when accelerated by a voltage of 10 kV, has a de-Broglie wavelength ' $\lambda$ '. If the voltage is increased to 20 kV then the deBroglie wavelength associated with the electron beam would be

An electron accelerated by a potential difference ' $V$ ' has de-Broglie wavelength ' $\lambda$ '. If the electron is accelerated by a potential difference ' 9 V ', its de-Broglie wavelength will be

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of ' $n$ ' is

A sub-atomic particle of mass $10^{-30}\text{ kg}$ is moving with a velocity $2.21 \times 10^{6}\text{ m/s}$. Under the matter wave consideration, the particle will behave closely like $(\mathrm{h} = 6.63 \times 10^{-34}\text{ J s})$

The de-Broglie wavelength of a neutron at $27^\circ\text{C}$ is ' $\lambda_0$ '. What will be its wavelength at $927^\circ\text{C}$?

The wavelength ' $\lambda$ ' of a photon and the deBroglie wavelength of an electron have same value. The ratio of kinetic energy of the electron to the energy of a photon is

( $\mathrm{m} =$ mass of electron, $\mathrm{c} =$ velocity of light, $\mathrm{h} =$ Planck's constant)

A light of wavelength $\lambda$ is incident on a photosensitive surface of negligible work function. The photoelectrons emitted from the surface have de-Broglie wavelength $\lambda_1$. Then ratio $\lambda : \lambda_1^{\,2}$ is ( $\mathrm{h} =$ Planck's constant, $\mathrm{c} =$ velocity of light, $\mathrm{m} =$ mass of electron)

A photon and an electron (mass $m$) have the same energy $E$. The ratio ( $\lambda$photon $/\lambda$electron) of their de Broglie wavelengths is: ( $c$ is the speed of light)

A proton of mass ' $m_P$ ' has same energy ' $E$ ' as that of a photon of wavelength ' $\lambda$ '. If the proton is moving at non-relativistic speed, then ratio of its de Broglie wavelength to the wavelength of photon is.

An electron of mass ' m ' with an initial velocity$\vec{v} = v_0\hat{i}\ (v_0 > 0)$ enters an electric field $\vec{E} = -E_o\hat{k}$. If the initial de Broglie wavelength is $\lambda_0$, the value after time t would be

When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number $= n$ ), the de-Broglie wavelength ( $\lambda$) associated with it is related to $n$ as

De-Broglie wavelength of an electron orbiting in the $n = 2$ state of hydrogen atom is close to (Given Bohr radius $= 0.052\text{ nm}$)

An electron in the ground state of the hydrogen atom has the orbital radius of $5.3 \times 10^{-11}\text{ m}$ while that for the electron in third excited state is $8.48 \times 10^{-10}\text{ m}$. The ratio of the de Broglie wavelengths of electron in the ground state to that in the excited state is

Let $E_c$ and $E_p$ represents kinetic energy of electron and photon respectively. If de-Broglie wavelength of a photon is twice the de-Broglie wavelength of an electron then $E_p/E_c$ is (speed of electron $= C/100$ where C is the velocity of light)

A photoemissive substance is illuminated with a radiation of wavelength $\lambda_i$ so that it releases electrons with de-Broglie wavelength $\lambda_e$. The longest wavelength of radiation that can emit photoelectron is $\lambda_o$. Expression for de-Broglie wavelength is given by:

(m: mass of the electron, h: Planck's constant and c: speed of light)

An electron of mass ' $m$ ' and charge ' $e$ ' initially at rest gets accelerated by a constant electric field ' E '. The rate of change of de-Broglie wavelength of the electron at time ' $t$ ' is

(Ignore relativistic effect)( $h =$ Planck's constant)

An electron with mass ' m ' with an initial velocity $(t = 0)\ \vec{v} = v_0\hat{i}\ (v_0 > 0)$ enters a magnetic field $\vec{B} = B_0\hat{j}$. If the initial de-Broglie wavelength at $t = 0$ is $\lambda_0$ then its value after time ' t ' would be:

An electron is released from rest near an infinite non-conducting sheet of uniform charge density '$-\sigma$'. The rate of change of de-Broglie wave length associated with the electron varies inversely as nth power of time. The numerical value of n is _____.

The kinetic energy of an electron is increased by 2 times, then the de-Broglie wavelength associated with it changes by a factor.

The graph which shows the variation of $\left(\frac{1}{\lambda^2}\right)$ and its kinetic energy, $E$ is (where $\lambda$ is de Broglie wavelength of a free particle):

Electrons are accelerated through a potential difference of 16 kV. If the potential difference is increased to 64 kV, then de-Broglie wavelength associated with electron will

If the potential difference used to accelerate electrons is doubled. By what factor does the deBroglie wavelength $(\lambda)$ associated with the electrons change?

If the potential difference used to accelerate electrons is increased four times, by what factor does the de-Broglie wavelength associated with the electrons change?

An electron and an alpha particle are accelerated by the same potential difference. Let $\lambda_e$ and $\lambda_\alpha$ denote the de-Broglie wavelengths of the electron and the alpha particle, respectively, then:

A proton, an electron and an alpha particle have the same energies. Their de-Broglie wavelengths will be compared as:

A proton and an electron have the same de Broglie wavelength. If $\mathrm{K_p}$ and $\mathrm{K_e}$ be the kinetic energies of proton and electron respectively, then choose the correct relation:

The de Broglie wavelengths of a proton and an $\alpha$ particle are $\lambda$ and $2\lambda$ respectively. The ratio of the velocities of proton and $\alpha$ particle will be:

A proton and an electron are associated with same de-Broglie wavelength. The ratio of their kinetic energies is:

(Assume h = 6.63 $\times 10^{-34}\text{ J s},\ \mathrm{m_e} = 9.0 \times 10^{-31}\text{ kg}$ and $\mathrm{m_p} = 1836$ times $\mathrm{m_e}$)

Kinetic energy of a proton is equal to energy $E$ of a photon. Let ' $\lambda_1$ ' be the de-Broglie wavelength of proton and ' $\lambda_2$ ' be the wavelength of photon. If $\left(\frac{\lambda_1}{\lambda_2}\right) \propto E^n$ then the value of ' $n$ ' is

The ratio of the wavelength of a photon of energy ' $E$ ' to that of the electron of same energy is ( $m =$ mass of an electron, $c =$ speed of light, $h =$ Planck's constant)

The de-Broglie wavelength of an electron is the same as that of a photon. If velocity of electron is $25\%$ of the velocity of light, then the ratio of K.E. of electron and K.E. of photon will be:

When the electron orbiting in hydrogen atom in its ground state moves to the third excited state, the de-Broglie wavelength associated with it

The de-Broglie wavelength $(\lambda)$ of a particle is related to its kinetic energy (E) as

The de Broglie wavelength of an electron having kinetic energy $\mathrm{E}$ is $\lambda$. If the kinetic energy of electron becomes $\frac{E}{4}$, then its de-Broglie wavelength will be:

When an electron is accelerated through a potential '$V$', the de-Broglie wavelength associated with it is '$4\lambda$'. When the accelerating potential is increased to $4\ \mathrm{V}$, its wavelength will be

If the potential difference used to accelerate electrons is doubled, by what factor does the de- Broglie wavelength associated with electrons change?

The de Broglie wavelength associated with an electron, accelerated by a potential difference of 81 V is given by:

Electron beam used in an electron microscope, when accelerated by a voltage of 20 kV, has a de- Broglie wavelength of $\lambda_0$. IF the voltage is increased to 40 kV, then the de-Broglie wavelength associated with the electron beam would be:

Proton $(\mathrm{P})$ and electron (e) will have same de-Broglie wavelength when the ratio of their momentum is (assume, $\mathrm{m_p} = 1849\, \mathrm{m_e}$ ):

The ratio of de-Broglie wavelength of an $\alpha$ particle and a proton accelerated from rest by the same potential is $\frac{1}{\sqrt{m}}$, the value of m is -

An $\alpha$-particle, a proton and an electron have the same kinetic energy. Which one of the following is correct in case of their de-Broglie wavelength:

The de Broglie wavelength of a molecule in a gas at room temperature (300 K) is $\lambda_1$. If the temperature of the gas is increased to 600 K, then the de Broglie wavelength of the same gas molecule becomes

Graph shows the variation of de-Broglie wavelength $(\lambda)$ versus $\frac{1}{\sqrt{V}}$ where '$V$' is the accelerating potential for four particles A, B, C, D carrying same charge but of masses $\mathrm{m_1, m_2, m_3, m_4}$. Which on represents a particle of largest mass?

An electron accelerated through a potential difference '$V_1$ ' has a de-Broglie wavelength '$\lambda$'. When the potential is changed to '$V_2$' its de-Broglie wavelength increases by $50\%$. The value of $\left(\frac{V_1}{V_2}\right)$ is

An electron accelerated through a potential difference $V_1$ has a de-Broglie wavelength of $\lambda$. When the potential is changed to $V_2$, its de-Broglie wavelength increases by $50\%$. The value of $\left(\frac{V_1}{V_2}\right)$ is equal to

A proton and an $\alpha$-particle are accelerated from rest by $2\text{ V}$ and $4\text{ V}$ potentials, respectively. The ratio of their de-Broglie wavelength is:

The ratio of the de-Broglie wavelengths of proton and electron having same Kinetic energy: (Assume $m_p = m_e \times 1849$)

The kinetic energy of an electron, $\alpha$-particle and a proton are given as $4\text{ K}, 2\text{ K}$ and $\text{K}$ respectively. The de-Broglie wavelength associated with electron $(\lambda \mathrm{e}), \alpha$-particle $(\lambda\alpha)$ and the proton $(\lambda p)$ are as follows:

A proton moving with one tenth of velocity of light has a certain de Broglie wavelength of $\lambda$. An alpha particle having certain kinetic energy has the same de-Broglie wavelength $\lambda$. The ratio of kinetic energy of proton and that of alpha particle is:

If the kinetic energy of a free electron doubles, it's de Broglie wavelength ($\lambda$) changes by a factor

The equation $\lambda = \frac{1.227}{x}\text{ nm}$ can be used to find the de-Brogli wavelength of an electron. In this equation $x$ stands for:

Where

$\mathrm{m} =$ mass of electron

$\mathrm{P} =$ momentum of electron

$\mathrm{K} =$ Kinetic energy of electron

$\mathrm{V} =$ Accelerating potential in volts for electron

The de Broglie wavelength associated with an electron, accelerated through a potential difference of $15056\text{ V}$, is:

An $\alpha$ particle and a proton are accelerated from rest through the same potential difference. The ratio of linear momenta acquired by above two particles will be:

An $\alpha$ particle and a carbon 12 atom has same kinetic energy K. The ratio of their de-Broglie wavelengths $(\lambda_\alpha : \lambda_{C12})$ is:

A proton, a neutron, an electron and an $\alpha$ particle have same energy. If $\lambda$p, $\lambda$n, $\lambda$e and $\lambda_\alpha$ are the de Broglie's wavelengths of proton, neutron, electron and $\alpha$ particle respectively, then choose the correct relation from the following: