JEE Mains · Chemistry · STD 11 - 2. structure of atom
The de Broglie wavelength \(\left( \lambda \right)\) associated with a photoelectron varies with the frequency \((v)\) of the incident radiation as, [\(v_0\) is threshold frequency]:
- A \(\lambda \propto \frac{1}{{\left( {v - {v_0}} \right)}}\)
- B \(\lambda \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{4}}}}}\)
- C \(\lambda \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{3}{2}}}}}\)
- D \(\lambda \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{2}}}}}\)
Answer & Solution
Correct Answer
(D) \(\lambda \propto \frac{1}{{{{\left( {v - {v_0}} \right)}^{\frac{1}{2}}}}}\)
Step-by-step Solution
Detailed explanation
\(\lambda = \frac{h}{{mv}}\) According to Einstein's theory of photoelectric effect: \(hv = h{v_0} + KE\) \(hv = h{v_0} + \frac{1}{2}m{v^2}\) \(2h(v - {v_0}) = m{v^2}\) \(\frac{{2h(v - {v_0})}}{m} = {v^2}\) \(v \propto {(v - {v_0})^{\frac{1}{2}}}\)…
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