Question

An electron in the hydrogen atom jumps from excited state 'n' to the ground state. The wavelength of the photoelectron illuminates a photo-sensitive material that has a work function- ω (in eV). The stopping potential of the photoelectron is found to be 10 eV. In a sample electrons fall from 'n' and it is found only two lines belong to the visible range. Find the value of work function in ω (in eV).​

Answer 1

Answer:

hc=eV0+ϕ0=10eV+2.75eV=12.75eV

But λhc=13.6[121−n21]eV⇒1−n21=13.612.75

⇒n21=0.0625⇒n2=62510000=16⇒

n=4

Explanation:

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