Question

The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV, and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in:

• A. X-ray region
• B. ultra-violet region
• C. infra-red region
• D. visible region

Answer 1

Answer: (B)

ultra-violet region

The stopping potential is 5V. The corresponding energy is $5\times 1.602\times {10}^{-19}=8.0\times {10}^{-19}J$ ------ (1)

The threshold energy ia 6.2 eV.
Convert the unit in J.
$6.2\times 1.602\times {10}^{-19}=9.95\times {10}^{-19}J$ --- (2)

The energy of the incident radiation is (1) + (2) or $17.95\times {10}^{-19}$ J.

The frequency of incident radiation is $\nu =\frac{E}{h}=\frac{17.95\times {10}^{-19}}{6.26\times {10}^{-34}}=2.7\times {10}^{15}$ Hz.

This belongs to ultraviolet region whose frequency range is $7.9\times {10}^{14}$ to $2\times {10}^{16}$.