Question

The work function of a metal is $1eV$. Light of wavelength $3000\mathring{A}$ is incident on this metal surface. The velocity of emitted photoelectrons will be

• A. $10m{s}^{-1}$
• B. $1\times 10^{3}m{s}^{-1}$
• C. $1\times 10^{4}m{s}^{-1}$
• D. $1\times 10^{6}m{s}^{-1}$

Answer 1

Answer: (D)

$E_{k\max }=\left(E_{photon}-W\right)eV$
$=\left(\frac{12.42\times 10^{-7}}{3\times 10^{-7}}-1\right)eV$
$=(4-1)eV$
$E_{k\max }=3eV=3\times 1.6\times 10^{-19}J$
$\frac{1}{2}m{v}_{\max }^2=3\times 1.6\times 10^{-19}$
$v_{\max }^2=\frac{3\times 2\times 1.6\times 10^{-19}}{9.1\times 10^{-31}}$
$v_{\max }^2=10^{12}$
$v_{\max }=10^{6}m{s}^{-1}$