The ground state energy of hydrogen atom is -13.6eV. What are P.E. and K.E. of electron in this state?

Correct Answer - B Kinetic energy of electron in ground state =-(Totoal energy in this state) `" " --(13.6 )eV=13.6 eV`

Correct Answer - D The ground state of hydrogen (n=1) is represented by K, the first excited state (n=2) is represented by L,the second excited state (n=3) is represent byb...

Correct Answer - A Here, `E=-13.6eV=-13.6xx1.6xx10^(-19)=-2.2xx10^(-18)J` `E=(-e^2)/(8piepsilon_0r)` `therefore` As orbital radius, `r=(-e^2)/(8piepsilon_0E)=(9xx10^(9)xx(1.6xx10^(-19))^2)/(2xx(2.2xx10^(-18)))=5.3xx10^(-11)m`.

Correct Answer - D Here, `r_1=5.30xx10^(-11)m,v_1=2.2xx10^6ms^(-1)` In the second excited state, `r_n=n^2r_1,v_n=(v_1)/(n)` `therefore r_2=4r_1=4xx5.30xx10^(-11)m=2.12xx10^(-10)m` and `v_2=(v_1)/(2)=(2.2xx10^(6))/(2)ms^(-1)=1.1xx10^(6)ms^(-1)`

Correct Answer - C Ionization energy `= (Rhc)Z^(2) = 9hcR` (Note : `E_(n) = - (RhcZ^(2))/(n^(2))`)

Correct Answer - 2 Kinetic energy of electron `= 13.6 [1-1/9] eV = 12.1 eV`.

Correct Answer - B `E_(1) = (hc)/(lambda) [(1)/(n_(1)^(2)) - (1)/(n_(2)^(2))]` For second excited state to first excited state `E_(1) = (hc)/(lambda) [1/4 - 1/9] rArr (hc)/(lambda) ((5)/(36))` For first excited state...

Correct Answer - A::C::D `122.4 = (13.6Z^(2))/(1^(2)) rArr Z = 3, 91.8 = 122.4[1-(1)/(4)]` So an electron of KE `91.8 eV` can transfer its energy to this atom.

Correct Answer - C Excitation upto `n=3` is required so that visible length is emitted upon de-excitation. So required energy `=13.6 (1-1/9)=12.1 eV`

Correct Answer - 3 `2^(nd)` excited state will be the `3^(rd)` energy level. `En=13.6/n^(2) eV` or `E=13.6/9=1.51 eV`.