In a hydrogen atom, an electron jumps from the second orbit to the first orbit.The region of the electromagnetic spectrum in which this line will fall :

Infrared

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U.V

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Visible light

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None of the above

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Solution

The correct option is B U.V
As we know,
v = 3.29 $\times$ 10 $^{15}$ [ $\frac{1}{1^{2}}$ - $\frac{1}{2^{2}}$ ]
= 2.47 $\times$ 10 $^{15}$ cps
$λ$ = $\frac{c}{v}$ $λ$ = 3 $\times$ 10 $^{8}$ /2.47 $\times$ 10 $^{15}$ m =1.215 $\times$ 10 $^{- 7}$ = 1.215 10 $^{3}$ A
This frequency belongs to UV region of EMR.

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In a hydrogen atom, an electron jumps from the second orbit to the first orbit.The region of the electromagnetic spectrum in which this line will fall :

The correct option is B U.VAs we know,v = 3.29 × 1015 [112 - 122] = 2.47 × 1015 cpsλ = cvλ = 3 × 108/2.47 × 1015 m =1.215 × 10−7 = 1.215 103 AThis frequency belongs to UV region of EMR.