Stopping potential for a photoelectric experiment is v₀ = 3.2 V for wavelength λ. If wavelength is doubled, the stopping potential becomes v′₀ = 0.7 V. Find the wavelength λ.

The correct option is 248 nm.

Step-by-step Explanation:

According to Einstein's photoelectric equation, the relationship between the stopping potential, the wavelength of the incident light, and the work function of the metal is given by:

$e{V}_0=\frac{hc}{\lambda }-\Phi$

where:
• $e$ is the elementary charge,
• $V_0$ is the stopping potential,
��� $h$ is Planck's constant,
• $c$ is the speed of light,
• $\lambda$ is the wavelength of the incident light, and
• $\Phi$ is the work function of the metal.

Let's write the equations for the two cases described in the problem:

Case 1: For wavelength $\lambda$, the stopping potential is $v_0=3.2\text{V}$.
$e\left(3.2\right)=\frac{hc}{\lambda }-\Phi$ --- (Equation 1)

Case 2: When the wavelength is doubled to $2\lambda$, the stopping potential becomes $v_0^{\prime }=0.7\text{V}$.
$e\left(0.7\right)=\frac{hc}{2\lambda }-\Phi$ --- (Equation 2)

To eliminate the work function $\Phi$, we subtract Equation 2 from Equation 1:

$e\left(3.2\right)-e\left(0.7\right)=\left(\frac{hc}{\lambda }-\Phi \right)-\left(\frac{hc}{2\lambda }-\Phi \right)$

$2.5\text{eV}=\frac{hc}{\lambda }-\frac{hc}{2\lambda }$

$2.5\text{eV}=\frac{hc}{2\lambda }$

Multiplying both sides by 2, we obtain:

$5.0\text{eV}=\frac{hc}{\lambda }$

Rearranging the equation to solve for the wavelength $\lambda$:

$\lambda =\frac{hc}{5.0\text{eV}}$

Using the standard value for the product of Planck's constant and the speed of light, $hc\approx 1240\text{eV}·\text{nm}$:

$\lambda =\frac{1240\text{eV}·\text{nm}}{5.0\text{eV}}$

$\lambda =248\text{nm}$

Thus, the wavelength of the incident light is 248 nm.