In a photoemissive cell with executing wavelength , the fastest e… - Vidyadip

MCQ

In a photoemissive cell with executing 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

A
$v(3 / 4)^{1 / 2}$
B
$v(4 / 3)^{1 / 2}$
✓
Less than $v(4 / 3)^{1 / 2}$
D
Greater than $v(4 / 3)^{1 / 2}$

✓

Answer

Correct option: C.

Less than $v(4 / 3)^{1 / 2}$

$ h v-W_0=\frac{1}{2} m v_{\max }^2 \Rightarrow \frac{h c}{\lambda}-\frac{h c}{\lambda_0}=\frac{1}{2} m v_{\max }^2 $
$ \Rightarrow h c\left(\frac{\lambda_0-\lambda}{\lambda \lambda_0}\right)=\frac{1}{2} m v_{\max }^2 \Rightarrow v_{\max }=\sqrt{\frac{2 hc}{m}\left(\frac{\lambda_0-\lambda}{\lambda \lambda_0}\right)}$
When wavelength is $\lambda$ and velocity is $v$, then$v=\sqrt{\frac{2 h c}{m}\left(\frac{\lambda_0-\lambda}{\lambda \lambda_0}\right)}$
When wavelength is $\frac{3 \lambda}{4}$ and velocity is $v^{\prime}$ then$v^{\prime}=\sqrt{\frac{2 h c}{m}\left[\frac{\lambda_0-(3 \lambda / 4)}{(3 \lambda / 4) \times \lambda_0}\right]}$
Divide equation (ii) by (i), we get
$ \frac{v^{\prime}}{v}=\sqrt{\frac{\left[\lambda_0-(3 \lambda / 4)\right]}{\frac{3}{4} \lambda \lambda_0 \times \frac{\lambda \lambda_0}{\lambda_0-\lambda}}} $
$ v^{\prime}=v\left(\frac{4}{3}\right)^{1 / 2}\sqrt{\frac{\left[\lambda_0-(3 \lambda / 4)\right]}{\lambda_0-\lambda}}$
$ \text { i.e. } v^{\prime}>v\left(\frac{4}{3}\right)^{1 / 2}$

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