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

[1998]

(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 $(4 / 3)^{1 / 2} \cdot v$

Show Answer

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 $v(\frac{4}{3})^{1 / 2}$.