Question

When a metal surface is illuminated by light of wavelengths 400 nm and 250 nm, the maximum velocities of the photoelectrons ejected are v and 2v respectively. The work function of the metal is - (h = Planck's constant, c = velocity of light in air)

• A. 2 hc × 10⁶ J
• B. 1.5 hc × 10⁶ J
• C. hc × 10⁶ J
• D. 0.5 hc × 10⁶ J

Answer 1

Answer: (D)

0.5 hc × 10⁶ J

Answer 2

$\frac { 1 } { 2 }$= K.E._max = $\frac { \mathrm { hc } } { \lambda }$ – W

$\frac { 1 } { 2 }$mv₁² = $\frac { \mathrm { hc } } { \lambda _ { 1 } }$ – W ….(1)

$\frac { 1 } { 2 }$ mv₂² = – W ….(2)

Ž $\left( \frac { \mathrm { v } _ { 1 } } { \mathrm { v } _ { 2 } } \right) ^ { 2 }$ =$\frac { \frac { \mathrm { hc } } { \lambda _ { 1 } } - \mathrm { W } } { \frac { \mathrm { hc } } { \lambda _ { 2 } } - \mathrm { W } }$= $\left( \frac { \mathrm { v } } { 2 \mathrm { v } } \right) ^ { 2 }$

$\frac { \mathrm { hc } } { \lambda _ { 2 } }$ – W = 4 $\left( \frac { \mathrm { hc } } { \lambda _ { 1 } } - \mathrm { W } \right)$

3W = $\frac { 4 \mathrm { hc } } { \lambda _ { 1 } }$ –

W = $\frac { \mathrm { hc } } { 3 }$ $\left[ \frac { 4 } { \lambda _ { 1 } } - \frac { 1 } { \lambda _ { 2 } } \right]$

= $\frac { \mathrm { hc } } { 3 }$ $\left[ \frac { 4 } { 400 \times 10 ^ { - 9 } } - \frac { 1 } { 250 \times 10 ^ { - 9 } } \right]$= $\frac { \mathrm { hc } } { 3 }$ × 10⁹ × $\frac { 150 } { 100 \times 250 }$

= 0.5 hc × 10⁶ J