Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength is changed to (3λ)/(4), the speed of the fastest emitted electron will be: (a) = v(4/3)1/2 (b) = v(3/4)1/2 (c) > v(4/3)1/2 (d) < v(4/3)1/2

Answer: (c). 1/2mv² = (hc)/(λ) - φ At (3λ)/(4): 1/2mv'² = 4/3·(hc)/(λ) - φ = 4/3((hc)/(λ) - φ) + φ/3 So 1/2mv'² > 4/3·1/2mv² ⇒ v' > v(4/3)1/2 So the answer is (c) > v(4/3)1/2.

Radiation of wavelength λ, is incident on a photocell. The fastest… | JEE Main 2016

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Photoelectric Effect: Einstein's Equation

Radiation of wavelength $\lambda,$ is incident on a photocell. The fastest emitted electron has speed $v.$ If the wavelength is changed to $\frac{3\lambda}{4},$ the speed of the fastest emitted electron will be:

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Answer: (c)

$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$
At $\frac{3\lambda}{4}$ : $\frac{1}{2}mv'^2 = \frac{4}{3}\cdot\frac{hc}{\lambda} - \phi = \frac{4}{3}\left(\frac{hc}{\lambda} - \phi\right) + \frac{\phi}{3}$
So $\frac{1}{2}mv'^2 > \frac{4}{3}\cdot\frac{1}{2}mv^2 \Rightarrow v' > v\left(\frac{4}{3}\right)^{1/2}$
So the answer is (c) $> v\left(\frac{4}{3}\right)^{1/2}$ .

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