Question

A photosensitive metallic surface is illuminated alternatively with lights of wavelength 3100 and 6200 A . It is observed that maximum speeds of the photoelectrons in two cases are in 2:1 . the work function of the metal is (hc =12400 eVA)

Answer 1

Answer:

The wavefunction of metal is 4.6 J.

Explanation:

Answer By

the failure

Answer 2

The work function of the metal is $\dfrac{4}{3}\ eV$ .

Given :

A photosensitive metallic surface is illuminated alternatively with lights of wavelength 3100 and 6200 A .

Also , maximum speeds of the photo electrons in two cases are in 2:1 .

We know , $K.E=h\dfrac{c}{\lambda}-E_o$   Here , $E_o$ is the work function of the metal .

Therefore , we know Kinetic Energy is directly proportional to the square of speed .

Therefore , ratio of kinetic energy is 4 : 1 .

So ,

$\dfrac{K.E_A}{K.E_B}=\dfrac{h\dfrac{c}{\lambda_A}-E_o}{h\dfrac{c}{\lambda_B}-E_o}\\\\\\\dfrac{4}{1}=\dfrac{\dfrac{12400}{3100}-E_o}{\dfrac{12400}{6200}-E_o}\\\\\\\dfrac{4}{1}=\dfrac{4-E_o}{2-E_o}\\\\E_o=\dfrac{4}{3}\ eV$

Hence , this is the required solution .

Learn More :

Work Function

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