If `13.6eV` energy is required to separate a hydrogen atom into a proton and an electron, then the orbital radius of electron in a hydrogen atom is

Correct Answer - B The transition from higher to first orbit causes appearance of Lyman line

Correct Answer - A::C Angular momentum `=(nh)/(2pi)=(3h)/(2pi)implies n=3` Also `r=n^(2)/2 a_(0) implies (9a_(0))/2=3^(2)/Z a_(0) implies Z=2` For de-excitation `1/lambda=Rz^(2) [1/n_(1)^(2)-1/n_(2)^(2)]=4R[1/n_(1)^(2)-1/n_(2)^(2)]` For `n=3` to `n=1` : `1/lambda=4R[1/1-1/9] implies lambda=9/(32R)` For `n=3` to...

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

Correct Answer - B `E_(H)=13.6eV` (inoisation energy ) `E_(He^(+))=13.6(Z)^(2)=(13.6)(Z)^(2)=54.4eV`

Correct Answer - d `k=(1)/(2)mv^(2)`.In electric field,`Y =(qEx^(2))/(2mv^(2))=(qEx^(2))/(4k)` The curvature of the trajectory is proportional to `qE//K` since q,E and K are same for btoth particles ,hence the trajectories will have...

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 - 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 - 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`.