Question

In a photoemissive cell with executing wavelength $\lambda $ , the fastest electron has speed $ \, v$ . If the exciting wavelength is changed to $\frac{3 \lambda }{4}$ , the speed of fastest emitted electron will be

• A. Greater than $v\left(\frac{4}{3}\right)^{\frac{1}{2}}$
• B. $v\left(\frac{3}{4}\right)^{\frac{1}{2}}$
• C. $\left(\frac{4}{3}\right)^{\frac{1}{2}}$
• D. Less than $v\left(\frac{4}{3}\right)^{\frac{1}{2}}$

Answer 1

Answer: (A)

According to photoelectric equation,
$hv-\omega _{0}=\frac{1}{2} \, mv_{m a x}^{2}$
$\frac{h c}{\lambda }-\frac{h c}{\lambda _{0}}=\frac{1}{2} \, \, mv_{m a x}^{2}$
$hc\left(\frac{\left(\lambda \right)_{0} - \lambda \, }{\lambda ⋅ \left(\lambda \right)_{0}}\right)=\frac{1}{2}mv_{m a x}^{2}$
$v_{m a x}=\sqrt{\frac{2 h c}{m} \, \left(\frac{\left(\lambda \right)_{0} - \lambda \, }{\lambda ⋅ \left(\lambda \right)_{0}}\right)}$
When wavelength is $\lambda $ and velocity v, then
$v=\sqrt{\frac{2 h c}{m} \, \left(\frac{\left(\lambda \right)_{0} - \lambda \, }{\lambda ⋅ \left(\lambda \right)_{0}}\right)}$ ..... (i)
When wavelength is $\frac{3 \lambda }{4}$ and velocity is $v′$ then
$v^{′}=\sqrt{\frac{2 h c}{m} \left[\frac{\left(\lambda \right)_{0} - \left(\frac{3 \lambda }{4}\right)}{\left(\frac{3 \lambda }{4}\right) \times \left(\lambda \right)_{0}}\right]}$ ...... (ii
Divide equation (ii) by equation (i), we get
$\frac{v^{′}}{v}=\sqrt{\frac{\lambda _{0} - \frac{3 \lambda }{4}}{\frac{3 \lambda \lambda _{0}}{4}} \times \frac{\lambda \lambda _{0}}{\lambda _{0} - \lambda }}$
$v^{′}=v\left(\frac{4}{3}\right)^{\frac{1}{2}}\sqrt{\frac{\left[\left(\lambda \right)_{0} - \left(\frac{3 \lambda }{4}\right)\right]}{\lambda \left(\lambda \right)_{0}}}=v^{′}>v\left(\frac{4}{3}\right)^{\frac{1}{2}}$