Question

When a metal surface is illuminated by light of wavelength \( \lambda \), the stopping potential is 8V. When the same surface is illuminated by light of wavelength \( 3\lambda \), the stopping potential is 2V. The threshold wavelength for this surface is:

• A. 5\( \lambda \)
• B. 3\( \lambda \)
• C. 9\( \lambda \)
• D. 4.5\( \lambda \)

Answer 1

The Correct Option is C

The photoelectric effect equation, \(E_k = h(f - f_0)\), is employed to determine the threshold wavelength of a metal surface exposed to light. This equation links the kinetic energy of emitted electrons (\(E_k\)) to the incident light's frequency (\(f\)) and the material's work function (\(hf_0\)).

The kinetic energy of ejected electrons is further defined by \(E_k = eV_s\), where \(V_s\) is the stopping potential and \(e\) is the elementary charge.

The relationship between light's frequency and wavelength is given by \(f = \frac{c}{\lambda}\), where \(c\) is the speed of light.

Substituting \(f = \frac{c}{\lambda}\) into the photoelectric equation yields \(E_k = h\frac{c}{\lambda} - hf_0\). With \(E_k = eV_s\), this becomes \(eV_s = h\frac{c}{\lambda} - hf_0\).

We consider two scenarios with different stopping potentials:

1. Incident wavelength \(\lambda\) with stopping potential 8V: \(8e = h\frac{c}{\lambda} - hf_0\)
2. Incident wavelength \(3\lambda\) with stopping potential 2V: \(2e = h\frac{c}{3\lambda} - hf_0\)

Rearranging these equations to solve for \(hf_0\):

From scenario 1: \(hf_0 = h\frac{c}{\lambda} - 8e\)

From scenario 2: \(hf_0 = h\frac{c}{3\lambda} - 2e\)

Equating the expressions for \(hf_0\):

\(h\frac{c}{\lambda} - 8e = h\frac{c}{3\lambda} - 2e\)

Isolating terms involving \(h\frac{c}{\lambda}\) and constants:

\(h\frac{c}{\lambda} - h\frac{c}{3\lambda} = 8e - 2e\)

\(h c \left( \frac{1}{\lambda} - \frac{1}{3\lambda} \right) = 6e\)

Simplifying the expression in the parenthesis:

\(h c \left( \frac{2}{3\lambda} \right) = 6e\)

Solving for \(h\frac{c}{\lambda}\):

\(h\frac{c}{\lambda} = 9e\)

Now, substitute this back into the rearranged equation from scenario 1 to find \(hf_0\):

\(hf_0 = 9e - 8e = e\)

The threshold frequency \(f_0\) is related to the threshold wavelength \(\lambda_0\) by \(f_0 = \frac{c}{\lambda_0}\). Therefore, \(hf_0 = h\frac{c}{\lambda_0}\).

Setting \(h\frac{c}{\lambda_0}\) equal to \(e\):

\(h\frac{c}{\lambda_0} = e\)

We also know from the simplified equation \(h\frac{c}{\lambda} = 9e\). We can express \(h\frac{c}{\lambda_0}\) in terms of this:

\(\frac{h\frac{c}{\lambda_0}}{h\frac{c}{\lambda}} = \frac{e}{9e}\)

\(\frac{\lambda}{\lambda_0} = \frac{1}{9}\)

\(\lambda_0 = 9\lambda\)

The threshold wavelength \(\lambda_0\) is found to be \(9\lambda\).