Question

When a metal plate is exposed to a monochromatic beam of light of wavelength 400 nm, a negative potential of 1.1 V is needed to stop the photo current. Find the threshold wavelength for the metal.

Answer 1

• Threshold wavelength for the metal plate is approximately 620 nm.

Given-

Wavelength of the light is = 400 nm = 400 × 10⁻⁹ m

Negative potential or threshold potential is = 1.1 V

Now we can apply the following formula-

hc/λ = hc/λ₀ + eV₀

where h is the Planks constant, c is speed of light, λ is wavelength of light, λ₀ is the threshold wavelength and V₀ is the threshold potential.

By putting the values in the above formula we get -

6.63 × 10⁻³⁴ × 3 × 10⁸ / 400 × 10⁻⁹ = 6.63 × 10⁻³⁴ × 3 × 10⁸ / λ₀ + 1.6 × 10⁻¹⁹ × 1.1

⇒ 4.97 = 19.89 × 10⁻26/ λ₀ + 1.76

⇒ λ₀ = 6.196 × 10⁻⁷ m ≈ 620 nm

Answer 2

Threshold Wavelength is 620 nm

Explanation:

$\lambda = 400\ nm = 400 \times 10^-^9 m$

$V_0 = 1.1 V$

$\frac {hc}{\lambda} = \frac {hc}{\lambda _0} + eV_0$

$\frac {6.63 \times 10^{-34} \times 3 \times 10^8}{400 \times 10^-^9}$

⇒ $\frac {6.63 \times 10^{-34} \times 3 \times 10^8}{\lambda _0} + 1.6 \times 10^{-19} \times 1.1$

⇒ $4.97 = \frac {19.89 \times 10^{-26}}{\lambda _0} + 1.76$

⇒ $\lambda _0 = \frac {19.89 \times 10^{-26}}{3.21}$

⇒ $6.196 \times 10^-^7 m = 620\ nm$