MHT CET · Physics · Dual Nature of Matter
When photons of energy hv fall on a photosensitive surface of work function \(E_0\), photoelectrons of maximum energy \(k\) are emitted. If the frequency of radiation is doubled the maximum kinetic energy will be equal to ( \(\mathrm{h}=\) Planck's constant)
- A k
- B 2 k
- C \(\mathrm{k}+\mathrm{E}_0\)
- D \(\mathrm{k}+\mathrm{h} v\)
Answer & Solution
Correct Answer
(D) \(\mathrm{k}+\mathrm{h} v\)
Step-by-step Solution
Detailed explanation
In photoelectric effect,
\(\text {Total energy } =K \cdot E \text { of electrons }+\) \(\text {Work function } \)
\( h v =k+w_0 \)
\( \therefore w_0 =h v-k\)
If frequency is doubled, let max. \(k \cdot E\) be \(k^{\prime}\)
\(\therefore h \times 2 v =k^{\prime}+w_0\)
\(\therefore 2 h v =k^{\prime}+(h v-k)\)
\(\therefore k^{\prime} =h v+k\)
\(\text {Total energy } =K \cdot E \text { of electrons }+\) \(\text {Work function } \)
\( h v =k+w_0 \)
\( \therefore w_0 =h v-k\)
If frequency is doubled, let max. \(k \cdot E\) be \(k^{\prime}\)
\(\therefore h \times 2 v =k^{\prime}+w_0\)
\(\therefore 2 h v =k^{\prime}+(h v-k)\)
\(\therefore k^{\prime} =h v+k\)
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