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
In ground state , kinetic energy `= 13.6 eV`, Potential energy `= -27.2 eV`
In first excited state, kinetic energy `= 3.4 eV`, Potential energy `= -...
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
`10= .^(n)C_(2) implies n=5`, then 5 orbits are involved upon coming to second excited state
so `n^(th)` excited state is `6^(th) [2^(nd), 3^(rd), 4^(th), 5^(th), 6^(th)]`
