The vapours of Hg absorb some electrons accelerated by a potential difference of 4.5 volts as a result of which light is emitted. If the full energy of single incident e− is supposed to be converted into light emitted by single Hg atom, then the wave number of the light is:
3.88×106m−1
3.98×106m−1
3.56×106m−1
3.63×106m−1
A
3.63×106m−1
B
3.98×106m−1
C
3.88×106m−1
D
3.56×106m−1
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Solution
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Wave number is given by ¯ν=1λ If the full energy of a single incident electron is supposed to be converted into light emitted by a single Hg atom then, hcλ=4.5eV ¯ν=1λ=4.5eVhc=4.5eV1242eV−nm=3.63×106m−1
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