Question

The maximum kinetic energy of the photoelectrons is found to be $6.63\times {10}^{-19}J$. When the metal is irradiated with a radiation of frequency $2\times {10}^{15}Hz$, the threshold frequency of the metal is about

• A. 1 × 10¹⁵ S^-1
• B. 2 × 10¹⁵ S^-1
• C. 3 × 10¹⁵ S^-1
• D. 105×10¹⁵ S^-1

Answer 1

The correct option is A

1 × 10¹⁵ S^-1

$K{E}_{max}=h(v-v_0)$

$v-v_0=\frac{K{E}_{max}}{h}=\frac{6.63\times {10}^{-19}J}{6.63\times {10}^{-34}Js}=1\times {10}^{15}{s}^{-1}$

$v_0=(v-1\times {10}^{15})s^{-1}$ = $(2\times {10}^{15}-1\times {10}^{15})s^{-1}$

= $1\times {10}^{15}{s}^{-1}$

It may be noted that $v_0$ cannot be equal to or greater than the frequency of the incident radiation ($2\times {10}^{15}{s}^{-1}$) in the present case because photoelectrons have net kinetic energy. Hence choices (B) and (C) are ruled out.