TS EAMCET · Physics · Dual Nature of Matter
Initially a photon of wavelength \(\lambda_1\) falls on photocathode and emits an electron of maximum energy \(E_1\). If the wavelength of the incident photon is changed to \(\lambda_2\), the maximum energy of the electron emitted becomes \(E_2\). Then value of \(h c\) ( \(h=\) Planck's constant, \(c=\) velocity of light) is
- A \(h c=\frac{\left(E_1+E_2\right) \lambda_1 \lambda_2}{\lambda_2-\lambda_1}\)
- B \(h c=\frac{E_1-E_2}{\lambda_2-\lambda_1} \cdot\left(\lambda_1 \lambda_2\right)\)
- C \(h c=\frac{\left(E_1-E_2\right)\left(\lambda_2-\lambda_1\right)}{\lambda_1 \lambda_2}\)
- D \(h c=\frac{\lambda_2-\lambda_1}{\lambda_1 \lambda_2 E_2} \cdot E_1\)
Answer & Solution
Correct Answer
(B) \(h c=\frac{E_1-E_2}{\lambda_2-\lambda_1} \cdot\left(\lambda_1 \lambda_2\right)\)
Step-by-step Solution
Detailed explanation
From equation of photoelectric effect, we have \[ \begin{aligned} & E_1=\frac{h c}{\lambda_1}-W \\ & E_2=\frac{h c}{\lambda_2}-W \end{aligned} \] where, \(W\) is work function. \[ \begin{aligned} & E_1+W=\frac{h c}{\lambda_1} \\ & E_2+W=\frac{h c}{\lambda_2} \end{aligned} \]…
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