JEE Mains · Physics · STD 12 - Dual nature of radiation and matter
- JEE Mains 2023
- Easy
- Numerical
The difference between threshold wavelengths for two metal surfaces \(A\) and \(B\) having work function \(\phi_A=9\,eV\) and \(\phi_{ B }=4.5\,eV\) in \(nm\) is:(Given, \(hc =1242\,eV\,nm\))
Pick an option to check your answer.
Answer & Solution
Correct Answer
(B) \(138\)
Step-by-step Solution
Detailed explanation
\(\lambda_{ A }=\left(\frac{1242}{9}\right)=138\,nm\)
\(\lambda_{ B }=\left(\frac{1242}{4.5}\right)=276\,nm\)
\(\lambda_{ B }-\lambda_{ A }=138\,nm\)
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