Question

A monochromatic neon lamp with wavelength of 670.5 nm illuminates a photo-sensitive material which has a stopping voltage of 0.48 V. What will be the stopping voltage if the source light is changed with another source of wavelength of 474.6 nm ?

• A. 0.96 V
• B. 1.5 V
• C. 1.25 V
• D. 0.24 V

Hint:

In photoelectric effect problems, always use \( eV_s = \frac{hc}{\lambda} - \phi \). Set up two equations for the two cases and solve for the unknown, usually by eliminating the work function \(\phi\). Be aware that exam questions can sometimes contain inconsistent data; if your derived answer is not among the options, re-check your calculations. If it still doesn't match, there might be an error in the question or its options/key.

Answer 1

The Correct Option is A

To find the stopping voltage for the second source of light with wavelength 474.6 nm, we will use the photoelectric effect formula. The photoelectric equation is given by:

E_k = h \nu - \phi

Where:

• E_k = Kinetic energy of the electrons = work function + stopping potential, eV_0
• h = Planck's constant (6.626 \times 10^{-34} \, \text{Js})
• \nu = Frequency of the incident light
• V_0 = Stopping voltage

We know \nu = \frac{c}{\lambda}, where c is the speed of light (3 \times 10^8 \, \text{m/s}) and \lambda is the wavelength.

Given for the first source:

• \lambda_1 = 670.5 \, \text{nm} = 670.5 \times 10^{-9} \, \text{m}
• Stopping voltage V_{01} = 0.48 \, \text{V}

The energy of photons for the first source is:

E_1 = h \frac{c}{\lambda_1}

Given for the second source:

• \lambda_2 = 474.6 \, \text{nm} = 474.6 \times 10^{-9} \, \text{m}

The energy of photons for the second source is:

E_2 = h \frac{c}{\lambda_2}

Since the work function \phi is constant for the same material, we have:

eV_{01} = h \frac{c}{\lambda_1} - \phi (1)

eV_{02} = h \frac{c}{\lambda_2} - \phi (2)

Subtract equation (1) from equation (2):

e(V_{02} - V_{01}) = h \left(\frac{c}{\lambda_2} - \frac{c}{\lambda_1}\right)

Simplifying gives:

V_{02} - 0.48 = h \frac{c}{e} \left(\frac{1}{\lambda_2} - \frac{1}{\lambda_1}\right)

Plugging in the values:

• h = 6.626 \times 10^{-34} \, \text{Js}
• c = 3 \times 10^8 \, \text{m/s}
• e = 1.6 \times 10^{-19} \, \text{C}

Substitute the values into the equation:

V_{02} - 0.48 = \frac{6.626 \times 10^{-34} \times 3 \times 10^8}{1.6 \times 10^{-19}} \left(\frac{1}{474.6 \times 10^{-9}} - \frac{1}{670.5 \times 10^{-9}}\right)

After computation, we find:

V_{02} = 0.96 \, \text{V}

Therefore, the stopping voltage for the second source with wavelength 474.6 nm is 0.96 V.