Question

A metal surface is illuminated by light of two different wavelengths 248 nm and 310 nm. The
maximum speeds of the photoelectrons corresponding to these wavelengths are ph and fly respectively,
If the ratio nearly : th= 4:1 and hc = 1240 eV nm, the work function of the metal is​

Answer 1

The work function of the metal is 3.9 eV.

Explanation:

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

Ratio u₁/u₂ = 4/1 [∴ u₁²/u₂² = 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 mu₁² = hc/μ₁ - w

1/2 mu₁² = 1240/248 - w

1/2 mu₁² = 5ev - w ...(1)

For light of 310 nm wavelengths:  

1/2 mu₂² = hc/μ₂ - w

1/2 mu₂² = 1240/310 - w

1/2 mu₂² = 4ev - w ...(2)

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

$\frac{1/2mu1^2}{1/2mu2^2}=\frac{5ev - w}{4ev - w}$

$\frac{u1^2}{u2^2} =\frac{5 - w}{4 - w}$

16 = $\frac{5-w}{4 - w}$

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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