The energy of the electron in the ground state of hydrogen atom is `-13.6 eV`. Find the kinetic energy and potential energy of electron in this state.

Given ,total energy of the electron `E=-4.6 EV` (i) Kinetic energy of electron K=-(Total energy E,) `rArr " " =-E=-(-4.6)=4.6 eV` (ii) de- Brogile wavelength `lambda_(d)=(h)/(sqrt(2 mK))=(6.6xx10^(-34))/(sqrt(2xx9.1xx10^(-31)xx4.6xx1.6xx10^(-19)))` `0.57xx10^(-9)m=0.57 nm`

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 - C Here the ground state energy , (E) `=-13.6eV` since, K.E. of the electron `=-13.6eV` `=13-6xx1.6xx10^(-19)J=21.76xx10^(-19)J=2.18xx10^(-18)J`.

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 - 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 - B `U = - (27.2eV)/(n^(2)) + C , k = (13.6eV)/(n^(2))` `E = - (13.6 eV)/(n^(2)) + C , DeltaE = 13.6eV((1)/(n^(2)) - (1)/(m^(2)))` here in these question...

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`