Dual Nature of Radiation and Matter - Result Question 55
59. In a photo-emissive cell, with exciting wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be
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(a) $(3 / 4)^{1 / 2} \cdot v$
(b) $(4 / 3)^{1 / 2} \cdot v$
(c) less than $(4 / 3)^{1 / 2} \cdot v$
(d) greater than $(4 / 3)^{1 / 2} \cdot v$
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Answer:
Correct Answer: 59. (d)
Solution:
(d) $\frac{1}{2} m v^{2}=\frac{h c}{\lambda}-W_0$ or $\frac{h c}{\lambda}=\frac{1}{2} m v^{2}+W_0$ and
$ \begin{aligned} \frac{1}{2} m v_1^{2} & =\frac{h c}{(3 \lambda / 4)}-W_0 \\ & =\frac{4}{3}(\frac{1}{2} m v^{2}+W_0)-W_0 \end{aligned} $
So, $v_1$ is greater than $v(\frac{4}{3})^{1 / 2}$.
