Question no 6 please
Q.6. The work function of caesium is 2.14 eV. Find (a) the threshold frequency for caesium and (b) the wavelength of the incident light if photocurrent is brought to zero by stopping potential of 0.60 V.
(Ans : 5.164 × 10 14 Hz, 4537 A O )

Dear Student,

ϕ=hν0ν0=ϕh=2.144.136*10-15=5.164*10-16 Hzhcλ=2.14+0.6eV4.136*10-15*3*108λ=2.14+0.6eVλ=4525 A°Regards

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