Question

Silver has a work function of \(4.7\,eV\). When ultraviolet light of wavelength \(180\,nm\) is incident upon it, potential of \(7.7\,V\) is required to stop photoelectrons reaching collector plate. The potential required to stop electrons when light of wavelength \(200\,nm\) is incident upon silver is

• A. 1.5 V
• B. 1.85 V
• C. 1.95 V
• D. 2.37 V

Hint:

Use \(E(eV)=\dfrac{1240}{\lambda(nm)}\). Then \(V_s = E - \phi\) in volts because \(1eV=e\times 1V\).

Answer 1

The Correct Option is A

Step 1: Use stopping potential relation.
\[ eV_s = \frac{hc}{\lambda} - \phi \] Step 2: For \(\lambda_1 = 180\,nm\).
\[ eV_{s1} = \frac{hc}{\lambda_1} - \phi \] Given \(V_{s1} = 7.7\,V\).
Step 3: For \(\lambda_2 = 200\,nm\).
\[ eV_{s2} = \frac{hc}{\lambda_2} - \phi \] Step 4: Subtract the equations to remove \(\phi\).
\[ e(V_{s1}-V_{s2}) = hc\left(\frac{1}{\lambda_1}-\frac{1}{\lambda_2}\right) \] Step 5: Use energy in eV form.
\[ E = \frac{1240}{\lambda(nm)}\ eV \] So,
\[ E_1 = \frac{1240}{180} \approx 6.89\ eV \] \[ E_2 = \frac{1240}{200} = 6.2\ eV \] Step 6: Find stopping potential for \(\lambda_2\).
\[ K_{max2} = E_2 - \phi = 6.2 - 4.7 = 1.5\ eV \Rightarrow V_{s2} = 1.5\ V \] Final Answer: \[ \boxed{1.5\ V} \]