Question

In a photoelectric experiment, the stopping potential for incident light of wavelength \(4000\ \text{\AA}\) is \(2\ \text{V}\). If the wavelength is changed to \(3000\ \text{\AA}\), the new stopping potential will be approximately \[ (\text{Use }h=4.14\times 10^{-15}\ \text{eV s},\ c=3\times 10^8\ \text{m/s}) \]

• A. \(2\ \text{V}\)
• B. \(3.03\ \text{V}\)
• C. \(4.14\ \text{V}\)
• D. \(1.5\ \text{V}\)

Hint:

For photoelectric effect, use \(E=\frac{1240}{\lambda(\text{nm})}\) eV and then apply \(E=\phi+V_s\).

Answer 1

The Correct Option is B

Concept: Einstein's photoelectric equation is: \[ \frac{hc}{\lambda}=\phi+eV_s \] In electron-volt form: \[ E=\phi+V_s \] where \(E\) is photon energy in eV and \(V_s\) is stopping potential in volts.

Step 1: For wavelength \(4000\ \text{\AA}\): \[ \lambda_1=4000\ \text{\AA}=400\ \text{nm} \] Photon energy: \[ E_1=\frac{1240}{400} \] \[ E_1=3.10\ \text{eV} \] Given stopping potential: \[ V_1=2\ \text{V} \]

Step 2: Find work function \(\phi\). \[ E_1=\phi+V_1 \] \[ 3.10=\phi+2 \] \[ \phi=1.10\ \text{eV} \]

Step 3: For wavelength \(3000\ \text{\AA}\): \[ \lambda_2=3000\ \text{\AA}=300\ \text{nm} \] Photon energy: \[ E_2=\frac{1240}{300} \] \[ E_2=4.13\ \text{eV} \]

Step 4: Find new stopping potential. \[ E_2=\phi+V_2 \] \[ 4.13=1.10+V_2 \] \[ V_2=3.03\ \text{V} \] Therefore, \[ \boxed{3.03\ \text{V}} \]