When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is _______ m.

$(h = 6.63 \times 10^{- 34} J s, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} m/s)$ [2026]

Question

When a light of a given wavelength falls on a metallic surface the stopping potential for photoelectrons is 3.2 V. If a second light having wavelength twice of first light is used, the stopping potential drops to 0.7 V. The wavelength of first light is _______ m.

$(h = 6.63 \times 10^{- 34} J s, e = 1.6 \times 10^{- 19} C, c = 3 \times 10^{8} m/s)$ [2026]

Smart Achievers · Accepted Answer

(4)

$q (3.2) = \frac{h c}{λ} - ϕ \dots (1)$

$q (0.7) = \frac{h c}{2 λ} - ϕ \dots (2)$

$Eq. (1) - Eq. (2)$

$q (2.5) = \frac{h c}{2 λ}$

$2.5 = (\frac{h c}{e}) (\frac{1}{2 λ})$

$2.5 = \frac{12400}{2 λ}$

$λ = \frac{12400}{5} Å$

$λ = 2480 Å$

$λ = 2.48 \times 10^{- 7} m$

Solution

1 $2.2 \times 10^{- 8}$

2 $3.1 \times 10^{- 7}$

3 $2.9 \times 10^{- 8}$

4 $2.5 \times 10^{- 7}$

Ans.
(4)

$q (3.2) = \frac{h c}{λ} - ϕ \dots (1)$

$q (0.7) = \frac{h c}{2 λ} - ϕ \dots (2)$

$Eq. (1) - Eq. (2)$

$q (2.5) = \frac{h c}{2 λ}$

$2.5 = (\frac{h c}{e}) (\frac{1}{2 λ})$

$2.5 = \frac{12400}{2 λ}$

$λ = \frac{12400}{5} Å$

$λ = 2480 Å$

$λ = 2.48 \times 10^{- 7} m$

Want Us to Email you About Special Offers & Updates?

Solution

1 2.2×10-8

2 3.1×10-7

3 2.9×10-8

4 2.5×10-7