Question

The de Broglie wavelength (\(\lambda\)) associated with a photoelectron varies with the frequency (\(\nu\)) of the incident radiation as, [\(\nu_0\) is threshold frequency] :

• A. \(\lambda \propto \frac{1}{(\nu-\nu_0)^{3/2}}\)
• B. \(\lambda \propto \frac{1}{(\nu-\nu_0)^{1/4}}\)
• C. \(\lambda \propto \frac{1}{(\nu-\nu_0)}\)
• D. \(\lambda \propto \frac{1}{(\nu-\nu_0)^{1/2}}\)

Hint:

Kinetic energy of the photoelectron is \(h(\nu-\nu_0)\), and \(\lambda = h/\sqrt{2mK}\).

Answer 1

The Correct Option is D

Einstein's equation: \(h\nu = h\nu_0 + \tfrac12 mv^2\), so \(\tfrac12 mv^2 = h(\nu-\nu_0)\).
Hence \(v = \sqrt{\frac{2h(\nu-\nu_0)}{m}}\).
de Broglie wavelength: \(\lambda = \frac{h}{mv} = \frac{h}{\sqrt{2mh(\nu-\nu_0)}}\).
Therefore \(\lambda \propto \frac{1}{(\nu-\nu_0)^{1/2}}\).
Answer: (4) \(\lambda \propto \frac{1}{(\nu-\nu_0)^{1/2}}\)