Chemistry, asked by madireddisanyasirao, 11 months ago

a light source of wavelength Lambda illuminates a metal and the photoelectrons with k.E one electron volts and another light source of wave lenghts of wave length lambda decided by three ,ejects photo electrons from some metal with k.E 4 ev find the value of work function​

Answers

Answered by physisit
0

Answer:

work function = hvo, where h is planck's constant and vo is the threshold energy

h = 6.626*10^-34

k.e is given in ev convert it into joule

4 ev = 4*1.602*10^-19

therefore work function = (6.626*10^-34)*4*1.602*10^-19

=42.4 *10^-53 J

Explanation:

hope it is the answer.

Answered by savithribathula1981
0

Answer:

E=KE+W

E is the total energy of the incident light source (E=hν=λhc)

KE is the kinetic energy of the ejected photoelectron

W is the work function of the metal. It is constant for a particular metal.

When light of wavelength λ is incident on a metal then,

λhc=1+W

When light of wavelength 3λ is incident

λ/3hc=5+W

λ3hc=5+W

Putting the value of λhc, we get

3(1+W)=5+W

2W=2

W=1eV

Explanation:

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