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QJEEM22Phf01e: JEE Main 2022 Physics - Class 12 JEEM22Phf01e Solution: In the hydrogen spectrum, be the wavelength of first transition …

In the hydrogen spectrum, λ\lambda be the wavelength of first transition line of Lyman series. The wavelength difference will be "a λ\lambda '' between the wavelength of 3rd 3^{\text {rd }} transition line of Paschen series and that of 2nd 2^{\text {nd }} transition line of Balmer series where a=\mathrm{a}= ___________.

Step-by-Step Verified Solution

1λ=RH(112−122){1 \over \lambda } = {R_H}\left( {{1 \over {{1^2}}} - {1 \over {{2^2}}}} \right)

1λ3=RH(132−162){1 \over {{\lambda _3}}} = {R_H}\left( {{1 \over {{3^2}}} - {1 \over {{6^2}}}} \right)

1λ2=RH(122−142){1 \over {{\lambda _2}}} = {R_H}\left( {{1 \over {{2^2}}} - {1 \over {{4^2}}}} \right)

∴\therefore λ3−λ2=aλ{\lambda _3} - {\lambda _2} = a\lambda

a=5a = 5

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