Which light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is 6.0 V. This potential drops to 0.6 V if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively ? [Take $\frac{hc}{e}=1.24\times {10}^{-6}{\text{J m C}}^{-1}$].

The correct option is 1.72 × 10^–7 m, 1.20 eV.

Step-by-step Explanation:

According to Einstein's photoelectric equation, the stopping potential $V_s$ is related to the wavelength of incident light $\lambda$ and the work function of the metal $\phi$ by the equation:

$e{V}_s=\frac{hc}{\lambda }-\phi$

Dividing the entire equation by the elementary charge $e$, we get the stopping potential in volts (V):

$V_s=\frac{hc}{e\lambda }-\frac{\phi }{e}$

Let ${\phi }_{\text{eV}}=\frac{\phi }{e}$ be the work function in electron-volts (eV), and let $K=\frac{hc}{e}=1.24\times {10}^{-6}\text{V m}$.

Case 1:
When light of wavelength $\lambda$ is used, the stopping potential is $6.0\text{V}$:

$6.0=\frac{K}{\lambda }-{\phi }_{\text{eV}}$    — (Equation 1)

Case 2:
When light of wavelength ${\lambda }^{\prime }=4\lambda$ is used, the stopping potential drops to $0.6\text{V}$ (Note: Intensity does not affect stopping potential):

$0.6=\frac{K}{4\lambda }-{\phi }_{\text{eV}}$    — (Equation 2)

Step 1: Finding the Work Function (${\phi }_{\text{eV}}$)

Multiply Equation 2 by 4:

$4\times 0.6=4\times \left(\frac{K}{4\lambda }-{\phi }_{\text{eV}}\right)$

$2.4=\frac{K}{\lambda }-4{\phi }_{\text{eV}}$    — (Equation 3)

Subtract Equation 3 from Equation 1:

$6.0-2.4=\left(\frac{K}{\lambda }-{\phi }_{\text{eV}}\right)-\left(\frac{K}{\lambda }-4{\phi }_{\text{eV}}\right)$

$3.6=3{\phi }_{\text{eV}}$

${\phi }_{\text{eV}}=\frac{3.6}{3}=1.20\text{eV}$

Step 2: Finding the Wavelength ($\lambda$)

Substitute ${\phi }_{\text{eV}}=1.20\text{eV}$ into Equation 1:

$6.0=\frac{K}{\lambda }-1.20$

$\frac{K}{\lambda }=7.20\text{V}$

$\lambda =\frac{K}{7.20}=\frac{1.24\times {10}^{-6}}{7.20}$

$\lambda \approx 0.1722\times {10}^{-6}\text{m}=1.72\times {10}^{-7}\text{m}$

Therefore, the wavelength of the first source and the work function of the metal are 1.72 × 10^–7 m and 1.20 eV, respectively.