Question

The de Broglie wavelength of the most energetic photoelectrons emitted from a photosensitive metal of work function \( \phi \), when light of frequency \( \nu \) is incident on it, is \( \lambda \). Then find \( \nu \) in terms of Planck’s constant \( h \), mass of electron \( m \), and other constants.  

• A. \( \frac{2\phi h}{m\lambda^2} \)
• B. \( \frac{2\phi h}{h m\lambda^2} \)
• C. \( \frac{\phi h}{h 2m\lambda^2} \)
• D. \( \frac{\phi h}{h 2m^2} \)

Hint:

The de Broglie wavelength of photoelectrons depends on the energy of the electron and the properties of the incident light. Ensure to use the right relations for momentum and energy when dealing with photoelectric effects.

Answer 1

The Correct Option is B

According to the photoelectric equation: \[ E_k = h\nu - \phi \] where \( E_k \) is the kinetic energy of the most energetic photoelectron, \( h \) is Planck's constant, and \( \phi \) is the work function of the metal. From de Broglie's equation for the wavelength of a particle, we have: \[ \lambda = \frac{h}{p} \] where \( p \) is the momentum of the electron, and momentum \( p = \sqrt{2mE_k} \). Substitute for \( E_k \) from the photoelectric equation: \[ \lambda = \frac{h}{\sqrt{2m(h\nu - \phi)}} \] Now, replacing \( \nu \) with \( \frac{c}{\lambda} \), you will get the final expression for the de Broglie wavelength in terms of \( \phi, h, m, \) and \( \lambda \), matching option (2). Thus, the correct answer is option (2), \( \frac{2\phi h}{h m\lambda^2} \).

Answer 2

Step 1: Given
- Work function of the metal = \( \phi \)
- Frequency of incident light = \( \nu \)
- De Broglie wavelength of the most energetic photoelectrons = \( \lambda \)

Step 2: Photoelectric equation
Energy of incident photon:
\[ E = h \nu \]
Maximum kinetic energy of photoelectron:
\[ K.E._{\max} = h \nu - \phi \]

Step 3: Kinetic energy in terms of velocity
\[ K.E._{\max} = \frac{1}{2} m v^2 \]
and the de Broglie wavelength is:
\[ \lambda = \frac{h}{m v} \implies v = \frac{h}{m \lambda} \]

Substitute \( v \) in kinetic energy:
\[ K.E._{\max} = \frac{1}{2} m \left(\frac{h}{m \lambda}\right)^2 = \frac{h^2}{2 m \lambda^2} \]

Step 4: Equate kinetic energies
\[ h \nu - \phi = \frac{h^2}{2 m \lambda^2} \implies h \nu = \phi + \frac{h^2}{2 m \lambda^2} \]

Step 5: Express \( \nu \)
\[ \nu = \frac{\phi}{h} + \frac{h}{2 m \lambda^2} \]

If the question wants \( \nu \) expressed only in terms of \( \phi, h, m, \lambda \), then the formula is:
\[ \nu = \frac{\phi}{h} + \frac{h}{2 m \lambda^2} \]

Note: The answer you gave, \( \frac{2 \phi h}{h m \lambda^2} \), is dimensionally incorrect and not consistent with the photoelectric equation.
The correct formula is as above.