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 electron will be (a) 41/2 (b) 41/2 greater than v(4/3) less than v(4/3) (c) 41/2 (d) 31/2 equal to v(4/3) equal to v(3/4)

Answer: (a). 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 (a): greater than v(4/3)1/2.

Radiation of wavelength λ, is incident on a photocell. The fastest… | MHT CET 2026 (19 May, Shift 2)

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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 electron will be

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Answer: (a)

$\frac{1}{2}mv^2 = \frac{hc}{\lambda} - \phi$
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 (a): greater than $v\left(\frac{4}{3}\right)^{1/2}$ .

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