Question

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

• A. $1.8\times {10}^{2}eV$
• 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 $4.37\times {10}^{-17}J$

Kinetic energy $=hv-h{v}_0=h(v-v_0)$ where $v_0$ is the threshold energy.

$\therefore \frac{v_2-v_0}{v_1-v_0}=\frac{K{E}_2/h}{K{E}_1/h}=1.5$

$\therefore 0.5v_0=1.5v_1-v_2=1.5(5.8\times {10}^{16})-8.1\times {10}^{16}=0.6\times {10}^{16}Hz$

$v_0=1.2\times {10}^{16}Hz$

The frequency of light with wavelength $3.846nm=\frac{3\times {10}^8}{3.846\times {10}^{-9}}=7.8\times {10}^{16}Hz$

$\therefore$ Energy of the photon emitted $=hv-h{v}_o=(7.8-1.2)\times {10}^{16}\times 6.63\times {10}^{-34}=4.37\times {10}^{-17}J$