The `H_(alpha)` line of Balmer series is obtained from the transitio n=3 (energy=-1.5eV) to n=2 (energy =-3.4 eV). What is the wavelength for this line. Given, `h=6.6xx10^(-34)Js:1eV=1.6xx10^(-19)J,c=3xx10^(8)ms^(-1)`

Options

A. 9210 Å
B. 8231 Å
C. 7321 Å
D. 6513 Å

Answer

D

Explanation

Here, `h=6.6xx10^(-34)J-s`
`1eV=1.6xx10^(-19)J,c=3xx10^(8)ms^(-1)`.
Energy of `n=3` level, `E_(3)=-1.5 eV`
Energy of n=2 level, `E_(2)=-3.4` eV
Energy of the emitted photon on transition from n=3 to n=2 level,
`hv=E_(3)-E_(2)=-1.5-(-3.4)=1.9eV`
or `(hc)/(lamda)=1.9eV=1.9xx1.6xx10^(-19)J`
or `lamda=(hc)/(1.9xx1.6xx10^(-19))=(6.6xx10^(-34)xx3xx10^(8))/(1.9xx1.6xx10^(-19))`
`=6513xx10^(-10)m=6513` Å