`H^(+)` ions are in excited state from where maximum 6 types of photons can be emitted. If E is the separation energy of electron in this excited state of `H^(+)` ion, then how many times of energy E is required to ionize Hydrogen atom ?
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 - A
(A)For `Li^(2+)`, n=6 to n=3
For H, the similar transition is 2 to 1
For `He^+` , the similar transition is 4 to 2
Energy of `4^(th)`...
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...