Question

The threshold frequency for a certain metal is n₀. When light of frequency n = 2n₀ is incident on it, the maximum velocity of photoelectrons is 4 × 10⁶ m/s. If the frequency of incident radiations is increased to 5n₀, then the maximum velocity of photoelectrons in m/s will be-

• A. $\frac{4}{5}$ × 10⁶
• B. 2 × 10⁶
• C. 8 × 10⁶
• D. 2 × 10⁷

Answer 1

Answer: (C)

8 × 10⁶

Answer 2

v_max. = $\sqrt { \frac { 2 } { \mathrm {~m} } \left( \mathrm { E } _ { \mathrm { Ph } } - \mathrm { W } \right) }$ =

(v_max.)₁ = $\sqrt { \frac { 2 h } { m } \left( 2 v _ { 0 } - v _ { 0 } \right) }$

(v_max)₂ = $\sqrt { \frac { 2 h } { m } \left( 5 v _ { 0 } - v _ { 0 } \right) }$

$\frac { \mathrm { v } _ { \max _ { .2 } } } { \mathrm { v } _ { \max _ { .1 } } }$ = 2 Ž $\mathbf { V } _ { \max _ { \cdot 2 } }$ =

$\mathbf { V } _ { \max _ { \cdot 2 } }$ = 2 × 4 × 10⁶ = 8 × 10⁶ m/s