Calculate the frequency of light emitted in an electron transition from the sixth to the second orbit of a hydrogen atom. In what region of the specturm does this frequency occur?
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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...
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In the nth orbit, the energy of an electron `E_(n)= -( 13.6)/(n^(2)) eV` for hydrogen atom. The energy rquired to take the electron from first orbit t
Correct Answer - A For n=2, using, `E_(n)=(-13.6eV)/(n^(2))E_(2)=-(13.6)/((2)^(2))=-3.4 eV` For n=1,`E_(1)=-13.6 eV` `:. " " E_(1rarr2)=-3.4-(-13.6)=10.2ev`
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The velocity of an electron in the first orbit of H-atom is v. The velocity of an electron in the 2nd orbit
Correct Answer - C `vprop(Z)/(n)` since Z and n both have become twice, velocity of electron in second orbit of `H^(+)` will be v.
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Assertion Second orbit circumference of hydrogen atom is two times the de-Broglie wavelength of electrons in that orbit Reason de-Broglie wavelength o
Correct Answer - B `mv_(2)r_(2)=2((h)/(2pi))` `:.2pir_(2)=2((h)/(mv_(2)))=2lambda_(2)` Further `lambda=(h)/(p)` Speed of momentum is maximum is ground state. Hence `lambda` is minimum
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Electrons are revolving around the nucleus in `n_(1^(th))` orbit of an atom, have atomic number `Z_1`, and in the `n_2` orbit of other atom, have atom
Correct Answer - A,B,D `L_1/L_2=(mv_1r_1)/(mv_2r_2)=(Z_1/n_1xxn_1^2/Z_1)/(Z_2/n_2xxn_2^2/Z_2)=n_1/n_2implies P_1/P_2(mv_1)/(mv_2)=(Z_1/n_1)/(Z_2/n_2)=(Z_1n_2)/(Z_2n_1)` `(f_1)/(f_2) =(v_1/(2pir_1))/(v_2/(2pir_2))=(Z_1/n_1xxZ_1/n_1^2)/(Z_2/n_2xxZ_2/n_2^2)=(Z_1/Z_2)^2.(n_2/n_1)^3implies (K.E._1)/(K.E._2)=(1/2mv_1^2)/(1/2mv_2^2)=(Z_1/Z_2)^2xx(n_2/n_1)^2=((Z_1n_2)/(Z_2n_1))^2` `(K.E_1)/(K.E_2)=(1/2(KZ_1e^2)/r_1)/(1/2(KZ_2e^2)/r_2)=(Z_1/n_1^2.Z_1)/(Z_2.Z_2/n_2^2)=(Z_1/Z_2.n_2/n_1)^2`
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The relation between the orbit radius and the electron velocity for a dynamically stable orbit in a hydrogen atom is (where, all notations have their
Correct Answer - C In hydrogen atom electrostatic force of attraction `(F_e)` between the revolving electrons and the nucleus provides the requisite centripetal force `(F_c)` to keep them in their orbits....
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the wavelength of radiation emitted is `lambda_0` when an electron jumps. From the third to second orbit of hydrogen atom. For the electron jumping fr
Correct Answer - B `(lambda)/(lambda_0)=([(1)/(2^2)-(1)/(3^2)])/([(1)/(2^2)-(1)/(4^2)])=(5)/(36)xx(16)/(3)=(20)/(27)`or `lambda=(20)/(27)lambda_0`
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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 - 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`.
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If a light with frequency `4xx10^(16) Hz` emitted photoelectrons with double the maximum kinetic as are emitted by the light of frequency `2.5xx10^(16
Correct Answer - `1xx10^(16) Hz` For frequency `2.5xx10^(16) Hz, h v=hv_(o)+KE_(max)` `:. h(2.5xx10^(16))=hv_(o)+KE_(max) ...(i)` For frequency `4xx10^(16)` Hz, `h(4xx10^(16))=hv_(o)+2KE_(max) ...(2)` Multiply `eq^(n)` (1) by `2` and substract `eq^(n)` (2) from it....
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If velocity of an electron in `I` orbit of `H` atom is `V`, what will be the velocity of electron in `3^(rd)` orbit of `Li^(2+)`
Correct Answer - A `V=2.188xx10^(6) Z/n m//s` Now, `V prop Z/n` so, `V_(Li^(2+))/V_(H)=-(Z_(1)//n_(1))/(Z_(2)//n_(2))=(3//3)/(1//1)=1` or, `V_(Li^(2+))=V_(H)`
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