Question

While studying photoelectric emission from a given surface, the wavelength of the incident radiation is changed from \(600\,\text{nm}\) to \(400\,\text{nm}\), keeping the intensity of radiation the same. Then :

• A. cut-off potential will decrease.
• B. cut-off potential will increase.
• C. saturation current will decrease.
• D. saturation current will increase.

Hint:

In photoelectric effect: \[ K_{\max}=h\nu-\phi \] and \[ eV_0=K_{\max}. \] Higher frequency (or lower wavelength) means higher stopping potential. Saturation current depends mainly on intensity.

Answer 1

The Correct Option is B

Concept: According to Einstein's photoelectric equation, \[ h\nu=\phi+K_{\max} \] where \[ h\nu=\text{energy of incident photon}, \] \[ \phi=\text{work function of the metal}, \] and \[ K_{\max}=\text{maximum kinetic energy of emitted electrons}. \] The stopping potential \(V_0\) is related to maximum kinetic energy by \[ eV_0=K_{\max}. \] Hence, a larger photon energy produces a larger stopping potential. Also, saturation current mainly depends on the intensity of incident light and not on its frequency.

Step 1: Examine the change in wavelength. The wavelength changes from \[ 600\,\text{nm} \] to \[ 400\,\text{nm}. \] Since \[ \nu=\frac{c}{\lambda}, \] a decrease in wavelength implies an increase in frequency. Thus, \[ \nu_{400}>\nu_{600}. \]

Step 2: Determine the effect on photon energy. Photon energy is \[ E=h\nu. \] Since frequency increases, \[ E_{400}>E_{600}. \] Therefore, each photon now carries more energy.

Step 3: Determine the effect on maximum kinetic energy. Using Einstein's equation, \[ K_{\max}=h\nu-\phi. \] Since \(h\nu\) increases while \(\phi\) remains constant, \[ K_{\max} \] increases.

Step 4: Determine the effect on stopping potential. Since \[ eV_0=K_{\max}, \] an increase in \(K_{\max}\) leads to an increase in stopping potential. Hence, \[ V_0 \] increases.

Step 5: Discuss saturation current. The intensity of radiation is kept constant. Saturation current depends primarily on the number of emitted photoelectrons per second, which in turn depends on intensity. Since intensity remains unchanged, \[ \text{Saturation Current} \] remains approximately unchanged. Thus options (C) and (D) are incorrect.

Step 6: Write the final conclusion. Reducing the wavelength from \(600\,\text{nm}\) to \(400\,\text{nm}\) increases the frequency, increases the photon energy, increases the maximum kinetic energy of photoelectrons, and hence increases the stopping potential. Therefore, \[ \boxed{\text{(B) cut-off potential will increase}} \] is the correct answer.