Question

For a photocell, the work function is \( \phi \) and the stopping potential is \( V_s \). The wavelength of the incident radiation can be expressed as

• A. \( \frac{hc}{e\phi + V_s} \)
• B. \( \frac{hc}{\phi + eV_s} \)
• C. \( \frac{hc}{\phi - eV_s} \)
• D. \( \frac{hc}{\phi} \)

Hint:

For photoelectric effect problems, use the photoelectric equation and rearrange to solve for the wavelength of the incident radiation.

Answer 1

The Correct Option is B

Step 1: Use of the photoelectric equation.
The photoelectric equation is given by: \[ \frac{hc}{\lambda} = \phi + eV_s \] where \( h \) is Planck's constant, \( c \) is the speed of light, \( \lambda \) is the wavelength, \( \phi \) is the work function, and \( V_s \) is the stopping potential.
Step 2: Solving for the wavelength.
Rearranging the equation to solve for \( \lambda \), we get: \[ \lambda = \frac{hc}{\phi + eV_s} \] Step 3: Conclusion.
Thus, the wavelength of the incident radiation is \( \frac{hc}{\phi + eV_s} \), corresponding to option (B).