a metal surface is illuminated by light of the two different wavelengths 248nm and 310nm the maximum speed of photoelectrons are u1 and u2 r

Answers (1)

Here, a metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm.

Ratio u?/u? = 4/1 [∴ u?²/u?² = 16/1]

hc = 1240 eV nm

 

=> According to the formula of kinetic energy:

KEmax = hc/λ - w

 

=> By substituting the value in above formula, we get

For light of 248 nm wavelengths:  

1/2 mu?² = hc/μ? - w

1/2 mu?² = 1240/248 - w

1/2 mu?² = 5ev - w ...(1)

 

For light of 310 nm wavelengths:  

1/2 mu?² = hc/μ? - w

1/2 mu?² = 1240/310 - w

1/2 mu?² = 4ev - w ...(2)

 

=> Dividing eq(1) by eq(2), we get

16 = 

16(4-w) = 5 - w

64 - 16w = 5 - w

64 - 5 = 16w - w

59 = 15w

w = 59/15

w = 3.93 eV ≈ 3.9 eV

 

Thus, the work function of the metal is 3.9 eV.

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