Question

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

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

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 } =3 \,eV =3 \times 1.6 \times 10^{-19} J $
$ \frac{1}{2} mv _{\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} $