Question

When a certain metal was irradiated using a light of frequency $8.1\times {10}^{16}$ Hz the photoelectron emitted had 1.5 times the kinetic energy as did the photoelectrons emitted when the same metal was irradiated using a light of frequency $5.8\times {10}^{16}$ Hz. If the same metal is irradiated using light of wavelength 3.846 nm, what will be the kinetic energy of the photoelectron emitted?

• A. 180 eV
• B. $3.65\times {10}^{-17}$ J
• C. $2.28\times {10}^2$ eV
• D. $4.37\times {10}^{-17}$ J

Answer 1

The correct option is D $4.37\times {10}^{-17}$ J

Given Information:
Frequency ${\nu }_1=5.8\times {10}^{16}$ Hz
Kinetic energy = $K.E_1$
Frequency ${\nu }_2=8.1\times {10}^{16}$ Hz
Kinetic energy = $1.5\times K.E_1$
work function = $h\nu -K.E$
so we get,
Work function = $h\times 5.8\times {10}^{16}-K.E_1$
Work function = $h\times 8.1\times {10}^{16}-1.5\times K.E_1$
On solving,
Work function $=7.944\times {10}^{-18}$ J
Kinetic energy $=3.045\times {10}^{-17}$ J
Wavelength of the incident radiation $=38.46$ $\stackrel{\circ }{A}$
Energy of the incident radiation = $h\nu$
$=\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{38.46\times {10}^{-10}}$
=$5.16\times {10}^{-17}$ J

Kinetic energy, $=5.16\times {10}^{-17}-7.94\times {10}^{-18}$ J
=$4.37\times {10}^{-17}$ J