Correct Answer - B
Energy , `E_n=-(13.6)/(n^2)eV`
In ground state energy `E_1=-(13-6)/(1^2)=-13.6eV`
In first excited state energy `E_2=-(13.6)/(2^2)=-3.4eV`
Then the required energy `=E_2-E_1=-3.4-(-13.6)=10.2eV`
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 - 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
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)]`
Correct Answer - B
Mole of `H_(2)` present in one litre `=(PV)/(RT)=(1xx1)/(0.0821xx298)=0.0409`
Thus, energy needed to break H-H bonds in `0.0409` mole `H_(2)=0.0409xx436=17.83 kJ`.
Also energy needed to excite one H...
