The photon emitted due to electronic transition of `5^(th)` excited state to `2^(nd)` excited state in `Li^(2+)`, is used to excite `He^+` already in

Monjur Alahi

The photon emitted due to electronic transition of `5^(th)` excited state to `2^(nd)` excited state in `Li^(2+)`, is used to excite `He^+` already in first excited state.`He^+` ion after absorbing the photon reaches in an orbit having total energy equal to :
A. `-3.4 eV`
B. `-13.6 eV`
C. `-6.8 eV`
D. `-27.2 eV`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

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)` orbit of `He^(+)` =`-13.6xx2^2/4^2= -3.4 eV`

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
Compare the general characteristics of the first transition series of transition metals with those of the second and third transition series metals in

(i) Electronif configuration. There are some exceptions in the electronic configurations. In all the three series. (ii) Oxidation state. The elements belonging to the different series but present in the...

2 Answers 1 views
Not considering the electronic spin, the degeneracy of the second excited state `(n = 3)` of H atom is 9, while the degeneracy of the second excited s

Correct Answer - A For `H^(-2)` nd excited state is `2p`, thus degeneracy is 3.

2 Answers 1 views
The energy of gamma `(gamma)` ray photon is `E_(gamma)` and that of an X-rays photon is `E_(X)`. If the visible light photon has an energy of `E_(v)`,

Correct Answer - C `E_(gamma) ge 100 keV, E_(X) = 100 eV` to `100 keV`. So, we can say that `E_(y) gt E_(x) gt E_(v)`,

2 Answers 1 views
In a hydrogen atom, a transition takes place from n = 3 to n = 2 orbit. Calculate the wavelength of the emitted photon. Will the photon be visible ? T

The wavelength of the emitted photon is given by `(1)/(lambda)=R((1)/n_(1)^(2)-(1)/(3^(2)))` When the transition takes place n =3 to n=2, then `(1)/(lambda)=(1.097xx10^(7))((1)/(2^(2))-(1)/(3^(3)))` `=1.097xx10^(7)xx(4)/(36)` `:." " lambda=(36)/(1.097xx10^(7)xx5)=6.563xx10^(-7) m` `=6563Å` `lambda` falsl in...

2 Answers 1 views
the energy required to excite an electron in hydrogen atom to its first excited state is

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`

2 Answers 1 views
A hydrogen atom initially in the ground level absorbs a photon and is excited to `n=4` level then the wavelength of photon is

Correct Answer - C Here ,`n_1=1` and `n_2=4` Energy of photon absorbed, `e=-e_2-e_1` `E_n=-(13.6)/n^2eV` Since, `E_2-E_1=-(13.6)/(4)^2-(-(13.6)/((1)^2))` `=-(13.6)/(16)+13.6=(13.6xx15)/(16)eV` `=12.75eV=12.75xx1.6xx10^(-19)J=20.4xx10^(-19)J` `E_2-E_1=(hc)/(lambda)` `therefore lambda =(hc)/(E_2-E_1)=(6.6xx10^(-34)xx3xx10^(8))/(20.4xx10^(10^(-19)))` `=9.70xx10^(-8)m=970xx10^(-10)=970Å`.

2 Answers 2 views
An electron in hydrogen atom first jumps from second excited state to first excited state and then from first excited state to ground state. Let the r

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...

2 Answers 1 views
A single electron ion has nuclear chrage `+Ze` where `Z` is atomic number and `e` is electronic charge. It requires `16.52 eV` to excite the electron

Correct Answer - (a) `Z=3`, (b) `108.8 eV`, (c) `1.013xx10^(-8) m`, (d) `122.4 eV` (a) `16.52=(E_(3)-E_(2))xxZ^(2)` `16.52=1.89xxZ^(2) rArr Z^(2)=9 "or" Z=3` (b) `(E_(3)-E_(1))_(z)=12.09xxZ^(2)=12.09xx9=108.81 eV` (c ) `lambda=12400/(DeltaE_(1 tp oo))=12400/(13.6xx3^(2))=101.3 Å=1.013xx10^(-8) m`,...

2 Answers 3 views
Find the quantum of the excited state of electrons in `He^(+)` ion which on transition to first excited state emit photons of wavelengths `108.5 nm`.

Correct Answer - B `1/lambda=R_(H)Z^(2)[1/1^(2)-1/n_(2)^(2)]` `1/(108.5xx10^(-9))=1.09678xx10^(7)xx4[1/2^(2)-1/n_(2)^(2)]` This gives `n_(2)=5`

2 Answers 1 views
`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 stat

Correct Answer - 4 `6=(n(n-1))/(2)" "n=4` Separation energy, `E=+13.6 xx(z^(2))/(n^(2))=+13.6xx((2)^(2))/((4)^(2))=3.4 eV` l.E. of Hydrogen atom `=x xxE=13.6 eV` `x=(13.6)/(E)=4`

2 Answers 2 views
X