Question

For a wavelength of 400 nm, kinetic energy
of emitted photoelectron is twice that for a
wavelength of 600 nm from a given metal.
The work function of metal is
[NCERT Pg. 395)
(1) 1.03 eV
(2) 2.11 eV
(3) 4.14 eV
(4) 2.43 eV​

Answer 1

refer to the attachment

4.13eV

HOPE IT HELPS YOU

Answer 2

Answer:

The work function of the metal is 1.03eV.i.e.option(1).

Explanation:

Let the kinetic energy associated with wavelength 600nm be "K". And with 400nm be "2K".

And from the photoelectric effect, we have,

$K=\frac{hc}{\lambda} -\phi$        (1)

Where,

K=kinetic energy associated with wavelength "λ"

h=planck's constant=6.626×10⁻³⁴Joule-second

c=speed of light in vacuum=3×10⁸m/s

Ф=work function of the metal

So, for solving this question we can write equation (1) as,

$\frac{2K}{K}=\frac{\frac{hc}{\lambda_1} -\phi}{\frac{hc}{\lambda_2} -\phi}$        (2)

From the question we have,

λ₁=400nm=400×10⁻⁹m

λ₂=600nm=600×10⁻⁹m

By substituting the values of wavelength in equation (2) we get;

$2=\frac{\frac{hc}{400\times 10^-^9} -\phi}{\frac{hc}{600\times 10^-^9} -\phi}$

$2(\frac{hc}{600\times 10^-^9}-\phi)=\frac{hc}{400\times 10^-^9}-\phi$

$\frac{2hc}{600\times 10^-^9}-2\phi=\frac{hc}{400\times 10^-^9}-\phi$

$hc(\frac{2}{600\times 10^-^9}-\frac{1}{400\times 10^-^9}) =2\phi-\phi$

$\phi=hc(\frac{4-3}{1200\times 10^-^9} )$

$\phi=6.626\times 10^-^3^4\times 3\times 10^8\times \frac{1}{1200\times 10^-^9}$

$\phi=1.6565\times 10^-^1^9 Joule$    (3)

We can convert joule into electron volt by dividing 1.6×10⁻¹⁹. So, equation (3) can be written as,

$\phi=\frac{1.6565\times 10^-^1^9 }{1.6\times 10^-^1^9} =1.03eV$

Hence, the work function of the metal is 1.03eV.i.e.option(1).