Question

The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from 1 to 2. establish a relation for threshold wavelength th in terms of 1 and 2?

Answer 1

AnswEr :-

Einstein's photoelectric equation states,

$\sf {h\nu=W_0 +KE}$

Where,

$\sf {\nu - Frequency\ of\ incident\ light}\\\\\sf {W_0 - Work\ potential\ of\ the\ metal}\\\\\sf {KE - Maximum\ kinetic\ energy\ of\ released\ electron}$

From given data,

$\sf {\frac{hc}{\lambda_1} = KE + W_o}\\\\\sf {\frac{hc}{\lambda_2} = 2KE + W_o}$

Eliminating KE, work function is given by:

$\sf {W_o = \frac{2hc}{\lambda_1} - \frac{hc}{\lambda_2}}$

Threshold wavelength is found by:

$\sf {\frac{hc}{\lambda_o} = W_o}$

Using above equations,

$\sf {\lambda_o = \frac{\lambda_1\lambda_2}{2\lambda_2-\lambda_1} }$

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Answer 2

The relation for threshold wavelength is $\dfrac{\lambda_{1}\lambda_{2}}{2\lambda_{2}-\lambda_{1}}$

Explanation:

Given that,

Initial wave length = λ₁

Final wavelength = λ₂

Suppose, Write Einstein's photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation

Einstein's photoelectric equation states,

$h\nu=W_{0}+(-KE)$

WHere $\nu$ =frequency of incident light,

$W_{0}$= the work potential of the metal

KE=maximum kinetc energy of released electron

(I). If $\nu<\nu_{0}$  

Then the maximum kinetic energy is negative, which is impossible.

(II). The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation.

We need to calculate the work function

Using photoelectric equation

$\dfrac{hc}{\lambda_{1}}=K.E+W_{0}$.......(I)

If the maximum kinetic energy of the photoelectrons gets double

$\dfrac{hc}{\lambda_{2}}=2K.E+W_{0}$...(II)

Subtract equation (II) from equation (I)

$\dfrac{2hc}{\lambda_{1}}-\dfrac{hc}{\lambda_{2}}=W_{0}$....(III)

We need to calculate the relation for threshold wavelength

Using formula of threshold wavelength

$\dfrac{hc}{\lambda_{0}}=W_{0}$

Put the value of threshold eavelength in equation (III)

$\dfrac{2hc}{\lambda_{1}}-\dfrac{hc}{\lambda_{2}}=\dfrac{hc}{\lambda_{0}}$....(III)

$\lambda_{0}=\dfrac{\lambda_{1}\lambda_{2}}{2\lambda_{2}-\lambda_{1}}$

Hence, The relation for threshold wavelength is $\dfrac{\lambda_{1}\lambda_{2}}{2\lambda_{2}-\lambda_{1}}$

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Topic : threshold wavelength

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