MHT CET · Physics · Dual Nature of Matter
Two identical photocathodes receive light of frequencies ' \(\mathrm{n}_1\) ' and ' \(\mathrm{n}_2\) '. If the velocities of the emitted photoelectrons of mass ' \(m\) ' are ' \(V_1\) ' and ' \(\mathrm{V}_2\) ' respectively, then ( \(\mathrm{h}=\) Planck's constant )
- A \(\mathrm{V}_1+\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)\right]^{1 / 2}\)
- B \(\mathrm{V}_1-\mathrm{V}_2=\left[\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)\right]^{1 / 2}\)
- C \(\mathrm{V}_1^2+\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1+\mathrm{n}_2\right)\)
- D \(\mathrm{V}_1^2-\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)\)
Answer & Solution
Correct Answer
(D) \(\mathrm{V}_1^2-\mathrm{V}_2^2=\frac{2 \mathrm{~h}}{\mathrm{~m}}\left(\mathrm{n}_1-\mathrm{n}_2\right)\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned}
& \text { K.E. } ._1=h n_1-\phi \\
& \text { K.E. } 2=\mathrm{hn}_2-\phi \\
& \therefore \quad \text { K.E. } 1 \text {-K.E. } 2=\text { hn }_1-\mathrm{hn}_2 \\
& \therefore \quad \frac{\mathrm{~m}\left(\mathrm{~V}_1^2-\mathrm{V}_2{ }_2\right)}{2}=\mathrm{h}\left(\mathrm{n}_1-\mathrm{n}_2\right) \\
& \therefore \quad \mathrm{V}_1^2-\mathrm{V}^2{ }_2=2 \mathrm{~h} \frac{\left(\mathrm{n}_1-\mathrm{n}_2\right)}{\mathrm{m}}
\end{aligned}\)
& \text { K.E. } ._1=h n_1-\phi \\
& \text { K.E. } 2=\mathrm{hn}_2-\phi \\
& \therefore \quad \text { K.E. } 1 \text {-K.E. } 2=\text { hn }_1-\mathrm{hn}_2 \\
& \therefore \quad \frac{\mathrm{~m}\left(\mathrm{~V}_1^2-\mathrm{V}_2{ }_2\right)}{2}=\mathrm{h}\left(\mathrm{n}_1-\mathrm{n}_2\right) \\
& \therefore \quad \mathrm{V}_1^2-\mathrm{V}^2{ }_2=2 \mathrm{~h} \frac{\left(\mathrm{n}_1-\mathrm{n}_2\right)}{\mathrm{m}}
\end{aligned}\)
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