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Q.

An electron passing through a potential difference of 4.9 V collides with a mercury atom and transfers it to the first excited state and stops. What is the wavelength of a photon corresponding to the transition of the mercury atom to its normal state ?

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a

2050 Ao

b

2240 Ao

c

2525 Ao

d

2935 Ao

answer is C.

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Detailed Solution

hcλ=E=eVλ=hceV6.6×1034×3×1081.6×1019×4.9=2525Ao

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