Whenever a photon is emitted by hydrogen in Balmar series, it is followed by another photon in Lyman series. The wavelength of latter photon will be

102 n m

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112 n m

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122 n m

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none of these

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Solution

The correct option is C $122 n m$
$λ = \frac{1240}{E (e V)} (n m) = \frac{1240}{10.2} = 122 n m$

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Whenever a photon is emitted by hydrogen in Balmar series, it is followed by another photon in Lyman series. The wavelength of latter photon will be

The correct option is C 122nm λ=1240E(eV)(nm)=124010.2=122 nm

The correct option is C $122 n m$
$λ = \frac{1240}{E (e V)} (n m) = \frac{1240}{10.2} = 122 n m$