a metal surface is illuminated by light of the two different wavelengths 248nm and 310nm the maximum speed of photoelectrons are u1 and u2 respectively if the ratio u1:u2 =2:1 & hxc=1240evnm the work function of the metal is nearly
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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