Solveeit Logo

Question

Physics Question on Photoelectric Effect

The work function of a metal is 1eV1\, eV. Light of wavelength 3000?3000 \,? is incident on this metal surface. The velocity of emitted photoelectrons will be

A

10ms110 \,ms ^{-1}

B

1×103ms11 \times 10^{3} ms ^{-1}

C

1×104ms11 \times 10^{4} ms ^{-1}

D

1×106ms11 \times 10^{6} ms ^{-1}

Answer

1×106ms11 \times 10^{6} ms ^{-1}

Explanation

Solution

Ekmax=(EphotonW)eVE _{ k \max } =\left( E _{ photon }- W \right) eV
=(12.42×1073×1071)eV=\left(\frac{12.42 \times 10^{-7}}{3 \times 10^{-7}}-1\right) eV
=(41)eV=(4-1) eV
Ekmax=3eV=3×1.6×1019JE _{ k \max } =3 \,eV =3 \times 1.6 \times 10^{-19} J
12mvmax2=3×1.6×1019\frac{1}{2} mv _{\max }^{2} =3 \times 1.6 \times 10^{-19}
vmax2=3×2×1.6×10199.1×1031v _{\max }^{2} =\frac{3 \times 2 \times 1.6 \times 10^{-19}}{9.1 \times 10^{-31}}
vmax2=1012v _{\max }^{2} =10^{12}
vmax=106ms1v _{\max } =10^{6} m s ^{-1}