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Atoms and Nuclei question

2011 · Q86
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Atoms and Nuclei question

2011 · Q86

NEETPhysicsAtoms and NucleiMCQ+4 / −1
An electron in the hydrogen atom jumps from excited state n to the ground state. The wavelength so emitted illuminates a photosensitive material having work function 2.75 eV. If the stopping potential of the photoelectron is 10 V, then the value of n is
  1. A
    2
  2. B
    3
  3. C
    4
  4. D
    5
View written solutionFree

Correct answer: C

KEmax = 10eV

ϕ\phi ϕ = 2.75 eV

Total incident energy

E = ϕ\phi ϕ + KEmax = 12.75 eV

∴\therefore∴ Energy is released when electron jumps from the excited state n to the ground state.

∴\therefore∴ E4 – E1 = {– 0.85 – (–13.6) ev} = 12.75eV

∴\therefore∴ value of n = 4

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