If the light of wavelength 412·5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why?
   

Metal Work Function (eV)
Na 1.92
K 2.15
Ca 3.20
Mo 4.17

 

 
 
 
 

Answers (1)
S safeer

Given,
      The wavelength of the incident light \lambda = 421\cdot 5 \; cm
The energy of the photon

. E= \frac{hc}{\lambda }= h\vartheta
We know 1eV= 1\cdot 60\times 10^{-19}J^
Therefore

E= \frac{hc}{\lambda \times 1\cdot 6\times 10^{-19}}J
So the energy of the photon is,
      E=\frac{\left ( 6\cdot 63\times 10^{-34} \right )\times \left ( 3\times 10^{8} \right )}{\left ( 412\cdot 5\times 10^{-9} \right )\times \left ( 1\cdot 6\times 10^{-19} \right )}\: eV
      E=3\cdot 01eV
The photoelectric emission takes place only when the energy of the incident photon is more than the work function of the metal.
So, From the give metal only Ca and Mo show the photoelectric emission.

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