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 - - 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 2 lines belong to the visible range- Find value of the work function - in eV

As only 2 lines in visible range are emitted so, the transitions are (4 → 2 and 3 → 2) nbsp; and n=4 Energy of photon, Δ E = 13.6 (1/1^2 1/4^2) eV = 12.75 ...

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Standard XII

Physics

Line Spectra

An electron i...

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 2 lines belong to the visible range. Find value of the work function $ω$ (in eV)

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Solution

As only 2 lines in visible range are emitted so, the transitions are $(4 \to 2 a n d 3 \to 2)$
and $n = 4$
Energy of photon,
$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) e V = 12.75 e V$
So,
$12.75 e V = ω +$ stopping potential $\Rightarrow 12.75 e V = ω + 10 e V$
$\Rightarrow ω = 2.75 e V$

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As only 2 lines in visible range are emitted so, the transitions are (4→2 and 3→2) and n=4 Energy of photon, ΔE=13.6(112−142) eV=12.75 eV So, 12.75 eV=ω+ stopping potential ⇒12.75 eV=ω+10 eV ⇒ω=2.75 eV

As only 2 lines in visible range are emitted so, the transitions are $(4 \to 2 a n d 3 \to 2)$
and $n = 4$
Energy of photon,
$Δ E = 13.6 (\frac{1}{1^{2}} - \frac{1}{4^{2}}) e V = 12.75 e V$
So,
$12.75 e V = ω +$ stopping potential $\Rightarrow 12.75 e V = ω + 10 e V$
$\Rightarrow ω = 2.75 e V$