Question

An electromagnetic wave of wavelength \(\lambda\) is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have de-Broglie wavelength \(\lambda _{1}\), then

• A. \(\lambda \propto \lambda _{1}\)
• B. \(\lambda \propto \lambda _{1}^{2}\)
• C. \(\lambda \propto \lambda _{1}^{4}\)
• D. \(\lambda \propto \lambda _{1}^{6}\)

Hint:

With negligible work function the kinetic energy equals the photon energy, then relate it to the de Broglie wavelength.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The photoelectric equation is \(\dfrac{hc}{\lambda}=\phi+K\). With negligible work function, \(K=\dfrac{hc}{\lambda}\).

Step 2: de Broglie Wavelength
The electron wavelength is \(\lambda_1=\dfrac{h}{\sqrt{2mK}}\).

Step 3: Combine
\(\lambda_1^2=\dfrac{h^2}{2mK}=\dfrac{h^2\lambda}{2m\,hc}=\dfrac{h\lambda}{2mc}\)

Step 4: Compare the options
Since \(h\), \(m\) and \(c\) are constants, \(\lambda\propto\lambda_1^2\), which is option (B).

Final Answer:
The incident wavelength is proportional to the square of the de Broglie wavelength, \(\lambda\propto\lambda_1^2\). \[ \boxed{\text{(B) }\lambda \propto \lambda _{1}^{2}} \]