An electron in a Bohr orbit of hydrogen atom with the quantum number `N_(2)` has an angular momentum `4.2176 xx 10^(-34) kg m^(2) s^(-1)`. If the electron drops from this level to the next lower level , the wavelength of this lines isA. `18 nm`B. `187.6 pm`C. `1876 Å`D. `1.876 xx 10^(4) Å`

An electron in a Bohr orbit of hydrogen atom with the quantum number `

1.	An electron in a Bohr orbit of hydrogen atom with the quantum number `N_(2)` has an angular momentum `4.2176 xx 10^(-34) kg m^(2) s^(-1)`. If the electron drops from this level to the next lower level , the wavelength of this lines isA. `18 nm`B. `187.6 pm`C. `1876 Å`D. `1.876 xx 10^(4) Å`
Answer» Correct Answer - D Angular momentum, `L = 4.2176 xx 10^(-34) = (n_(2) h)/(2 pi)` `implies n_(2) = 4` For the transtion from `n_(2) = 4 to n_(1) = 3`, the wavelength spectral line`= lambda` `(1)/(lambda) = (13.6)/(hc) ((1)/(3^(2)) - (1)/(4^(2)))` `= (13.6 eV)/(1240 eV nm) ((7)/(9 xx 16))` `lambda = (1240 xx 144)/(13.6 xx7) = 1876 nm = 18760 Å` `= 1.876 xx 10^(4) Å`

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An electron in a Bohr orbit of hydrogen atom with the quantum number `N_(2)` has an angular momentum `4.2176 xx 10^(-34) kg m^(2) s^(-1)`. If the electron drops from this level to the next lower level , the wavelength of this lines isA. `18 nm`B. `187.6 pm`C. `1876 Å`D. `1.876 xx 10^(4) Å`

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