Question

The maximum kinetic energy of the photoelectrons is found to be 6.63 × 10^–19 J when the metal is irradiated with a radiation of frequency 3 × 10¹⁵ Hz. The threshold frequency of the metal is –

• A. 1 × 10¹⁵ Hz
• B. 3 × 10¹⁵ Hz
• C. 2 × 10¹⁶ Hz
• D. 2 × 10¹⁵ Hz

Answer 1

Answer: (D)

2 × 10¹⁵ Hz

Answer 2

K.E. = hv – hv₀ = h(v – v₀)

v – v₀ = $\frac{K.E.}{h}$ = $\frac{6.63 \times 10^{–10}}{6.62 \times 10^{–34}}$ = 1 × 10¹⁵

v₀ = v – 1 × 10¹⁵ = 3 × 10¹⁵ – 1 × 10¹⁵

= 2 × 10¹⁵ Hz.