Question

In a photoemissive cell with exciting wavelength $\lambda$, the fastest electron has speed $v$. If the exciting wavelength is changed by $3\lambda /4$, the speed of the fastest emitted electron will be :

• A. $v{\left(3/4\right)}^{1/2}$
• B. $v{\left(4/3\right)}^{1/2}$
• C. less than $v{\left(4/3\right)}^{1/2}$
• D. greater than $v{\left(4/3\right)}^{1/2}$

Answer 1

Answer: (D)

From $E=W_0+\frac{1}{2}m{v}_{max}^2\Rightarrow v_{max}=\sqrt{\frac{2E}{m}-\frac{2W_0}{m}}$

where $E=\frac{hc}{\lambda }$

If wavelength of incident light charges from $\lambda$ to $\frac{3\lambda }{4}$ (Decreases)

Let energy of incident light charges from $E$ and speed of fastest electron changes from $v$ to $v^{\prime }$ then

$v^{\prime }=\sqrt{\frac{2E^{\prime }}{m}-\frac{2W_0}{m}}$

As $E\propto \frac{1}{\lambda }\Rightarrow E^{\prime }\frac{4}{3}E$

Hence $v^{\prime }=\sqrt{\frac{2\left(\frac{4}{3}\right)E}{m}-\frac{2W_0}{m}}$

$\Rightarrow v^{\prime }={\left(4/3\right)}^{1/2}\sqrt{\frac{2E}{m}-\frac{2W_0}{m{\left(4/3\right)}^{1/2}}}$

So, $v^{\prime }>{\left(4/3\right)}^{1/2}v$