In a photo-emissive cell, with exciting wavelength λ, the fastest electron has speed v. If the exciting wavelength is changed to 3λ/4, the speed of the fastest emitted electron will be (a) v(3/4)1/2 (b) v(4/3)1/2 (c) less than v(4/3)1/2 (d) greater than v(4/3)1/2

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

In a photo-emissive cell, with exciting wavelength λ, the fastest… | NEET 1998 (AIPMT)

Q171NEET 1998 · AIPMTMedium

Photoelectric Effect: Einstein's Equation

In a photo-emissive cell, with exciting wavelength $\lambda$ , the fastest electron has speed $v$ . If the exciting wavelength is changed to $3\lambda/4$ , the speed of the fastest emitted electron will be

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Answer: (d)

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

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