Question

The surface of a certain metal is first illuminated with light of wavelength \(\lambda_1=350\,\text{nm}\), and then by light of wavelength \(\lambda_2=540\,\text{nm}\). It is found that the maximum speed of the photoelectrons in the two cases differ by a factor of \(2\). The work function of the metal (in eV) is close to \[ \left(\text{Energy of photon}=\frac{1240}{\lambda(\text{nm})}\text{ eV}\right) \]

• A. \(2.58\)
• B. \(1.88\)
• C. \(3.22\)
• D. \(1.48\)

Hint:

If photoelectron speeds are related by \[ v_1=n\,v_2 \] then \[ K_1=n^2K_2 \] Use Einstein's equation \[ K_{\max}=h\nu-\phi \] and solve for the work function.

Answer 1

The Correct Option is B

Step 1:Calculate photon energies. For \[ \lambda_1=350\,\text{nm} \] \[ E_1=\frac{1240}{350} \] \[ E_1=3.543\,\text{eV} \] For \[ \lambda_2=540\,\text{nm} \] \[ E_2=\frac{1240}{540} \] \[ E_2=2.296\,\text{eV} \]

Step 2: Apply the condition \(K_1=4K_2\). \[ E_1-\phi = 4(E_2-\phi) \] Substituting values, \[ 3.543-\phi = 4(2.296-\phi) \] \[ 3.543-\phi = 9.184-4\phi \] \[ 3\phi = 9.184-3.543 \] \[ 3\phi = 5.641 \] \[ \phi = 1.88\,\text{eV} \]

Step 3: State the answer. \[ { \phi = 1.88\,\text{eV} } \] Hence, the correct option is \[ {(B)} \]