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 (Given $h=4.14\times 10^{-15}eVs$ and $c=3\times 10^{8}m{s}^{-1}$)

• A. $\approx 0.6\times 10^{6}m{s}^{-1}$
• B. $\approx 61\times 10^{3}m{s}^{-1}$
• C. $\approx 0.3\times 10^{6}m{s}^{-1}$
• D. $\approx 6\times 10^{5}m{s}^{-1}$

Answer 1

Answer: (A), (D)

${\lambda }_0=3250\times 10^{-10}m$
$\lambda =2536\times 10^{-10}m$
$\varphi =\frac{1242eV-nm}{325nm}=3.82eV$
$hv=\frac{1242eV-nm}{253.6nm}=4.89eV$
$K{E}_{\max }=(4.89-3.82)eV=1.077eV$
$\frac{1}{2}m{v}^2=1.077\times 1.6\times 10^{-19}$
$v=\sqrt{\frac{2\times 1.077\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}}$
$v=0.6\times 10^{6}m/s$