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when a certain energy is applied to an hydrogen atom, an electron jumps from n =1 to n = 3 state. Find (i) the energy absorbed by the electron. (ii) wavelength of radiation emitted when the electron jump back to its initial state. ( Energy of electron in first orbit = - 13.6 eV , Planck's constant = `6.6225 xx10^(-34)` Js, Charge on electron = `1.6 xx10^(-19)` C, speed of light in vacuum `= 3xx10^(8)ms^(-1)`

Text Solution

Verified by Experts

Given: `E_(1)=-13.6,E_(3)=(-13.6)/(3^(2))=-1.511eV`
Energy transition `E_(3)-E_(1)=(1.511-(-13.6))eV=12.089eV`
Also , `E=hv and v=(c)/(lamda)`
`:.lamda=(hc)/(E)=(6.625xx10^(-34)xx3xx10^(8))/(19.34xx10^(-19))m`
i.e., `lamda=1.0277xx10^(-7)m " or " lamda=1027.7Å`
wav number , `bar(v)=1/lamda=1/(1.0277xx10^(7))m^(-1) =(=0.97304466xx10^(7)m^(-1)`
i.e., `bar(v) = 973046.6m^(-1)`
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