On a surface, if photon of wavelength $\lambda$ is incident, the stopping potential is $3.2$ V. If the wavelength incident is $2\lambda$, stopping potential is $0.7$ V. Find $\lambda$.

Question

Light is incident on a metallic plate having work function $110 \times 10^{-20}\,\text{J}$. If the produced photoelectrons have zero kinetic energy, then the angular frequency of the incident light is _________ rad/s. ($h = 6.63 \times 10^{-34}\,\text{J·s}$)}

Answer 1

To solve this problem, we will apply concepts from the photoelectric effect, specifically using the photoelectric equation. The equation relating stopping potential and wavelength of incident light is:

$eV_s = \frac{hc}{\lambda} - \phi$

Where:

$e$: Charge of an electron ($1.6 \times 10^{-19} \, \text{C}$)
$V_s$: Stopping potential
$h$: Planck’s constant ($6.626 \times 10^{-34} \, \text{Js}$)
$c$: Speed of light ($3 \times 10^{8} \, \text{m/s}$)
$\lambda$: Wavelength of incident light
$\phi$: Work function of the surface

Given:

For wavelength $\lambda$, stopping potential $V_{s1} = 3.2 \, \text{V}$
For wavelength $2\lambda$, stopping potential $V_{s2} = 0.7 \, \text{V}$

Applying the equation for both scenarios, we get two equations:

$e \times 3.2 = \frac{hc}{\lambda} - \phi$
$e \times 0.7 = \frac{hc}{2\lambda} - \phi$

Substitute the known values for $e$, $h$, and $c$:

$1.6 \times 10^{-19} \times 3.2 = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi$
$1.6 \times 10^{-19} \times 0.7 = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{2\lambda} - \phi$

Let's solve these equations simultaneously:

From equation (1):

$\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{\lambda} - \phi = 5.12 \times 10^{-19}$

From equation (2):

$\frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{2\lambda} - \phi = 1.12 \times 10^{-19}$

Eliminate $\phi$:

$\frac{hc}{\lambda} - \frac{hc}{2\lambda} = 5.12 \times 10^{-19} - 1.12 \times 10^{-19}$

Simplifying:

$\frac{hc}{2\lambda} = 4 \times 10^{-19}$

Solving for $\lambda$:

$\lambda = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{8 \times 10^{-19}}$

Simplifying gives:

$\lambda = 2.48 \times 10^{-7} \, \text{m}$

Therefore, the correct answer is $2.48 \times 10^{-7} \, \text{m}$.

On a surface, if photon of wavelength $\lambda$ is incident, the stopping potential is $3.2$ V. If the wavelength incident is $2\lambda$, stopping potential is $0.7$ V. Find $\lambda$.

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The Correct Option is D

Solution and Explanation

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