Question

The photoelectric threshold wavelength of silver is $3250\times {10}^{-10}m$. The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $2536\times {10}^{-10}m$ is

• A. $0.6\times {10}^{6}m/s$
• B. $61\times {10}^{3}m/s$
• C. $0.3\times {10}^{6}m/s$
• D. $8\times {10}^{8}m/s$

Answer 1

The correct option is A $0.6\times {10}^{6}m/s$

The maximum kinetic energy is given as,
$K_{max}=hv-{\varphi }_0=hv-h{v}_0=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}\mathrm{where}{\lambda }_0=\mathrm{threshold}\mathrm{wavelength}or\frac{1}{2}m{v}^2=\frac{hc}{\lambda }-\frac{hc}{{\lambda }_0}$
Here,
$h=4.14\times {10}^{-15}eV,c=3\times {10}^{8}m/s{\lambda }_0=3250\times {10}^{-10}m=3250\stackrel{0}{A}\lambda =2536\times {10}^{-10}m=2536\stackrel{0}{A,}m=9.1\times {10}^{-31}Kg$
So,
$hc=(4.14\times {10}^{-15})\times (3\times {10}^8)=12420eV\stackrel{0}{A}\therefore \frac{1}{2}m{v}^2=12420\left(\frac{1}{2536}-\frac{1}{3250}\right]eV=1.076eV{v}^2=\frac{2.152eV}{m}=\frac{2.152\times 1.6\times {10}^{-19}}{9.1\times {10}^{-31}}\therefore v\approx 0.6\times {10}^{6}m/s$