JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2026
- Medium
- MCQ
\(K_1\) and \(K_2\) be the maximum kinetic energies of photoelectrons emitted from a surface of a given material for the light of wavelength \(\lambda_1\) and \(\lambda_2\), respectively. If \(\lambda_1=2\lambda_2\) then the work function of material is given by:
Pick an option to check your answer.
Answer & Solution
Correct Answer
(D) \(K_2-2K_1\)
Step-by-step Solution
Detailed explanation
Using Einstein's photoelectric equation:
\(K_1 = \dfrac{hc}{\lambda_1} - \phi\)
\(K_2 = \dfrac{hc}{\lambda_2} - \phi\)
Given \(\lambda_1 = 2\lambda_2\), the first equation becomes:
\(K_1 = \dfrac{hc}{2\lambda_2} - \phi\)
Multiplying by \(2\):
\(2K_1 = \dfrac{hc}{\lambda_2} - 2\phi\)
Substituting \(\dfrac{hc}{\lambda_2} = K_2 + \phi\) from the second equation:
\(2K_1 = (K_2 + \phi) - 2\phi\)
\(2K_1 = K_2 - \phi\)
\(\phi = K_2 - 2K_1\)
Answer: \(K_2-2K_1\)
\(K_1 = \dfrac{hc}{\lambda_1} - \phi\)
\(K_2 = \dfrac{hc}{\lambda_2} - \phi\)
Given \(\lambda_1 = 2\lambda_2\), the first equation becomes:
\(K_1 = \dfrac{hc}{2\lambda_2} - \phi\)
Multiplying by \(2\):
\(2K_1 = \dfrac{hc}{\lambda_2} - 2\phi\)
Substituting \(\dfrac{hc}{\lambda_2} = K_2 + \phi\) from the second equation:
\(2K_1 = (K_2 + \phi) - 2\phi\)
\(2K_1 = K_2 - \phi\)
\(\phi = K_2 - 2K_1\)
Answer: \(K_2-2K_1\)
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