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If the electron in H atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by
We know that
1/λ=R(1/n^{2}_{1}−1/n^{2}_{2})
1/λ=R(1/2^{2}) − (1/3^{2})⇒R(1/4)−(1/9)
1/λ=(9−4/36)R=5R/36⇒λ=36/5R
The ratio of the speed of the electron in the ground state of hydrogen atom to the speed of light is
The speed of revolving electron in nth state of hydrogen atom is:
v=e^{2}/2nhϵ_{0}
For n=1,
v= (1.6×10^{−19})^{2}/2(1)(6.6×10^{−34})(8.85×10^{−12})
v=2.56×10^{−38}/116.82×10^{−46}
v=0.0219×10^{8}ms^{−1}
The speed of light is 3×10^{8}
Hence,
v/c=0.0219×10^{8}/3×10^{8}
v/c=1/137
In hydrogen atom the kinetic energy of electron in an orbit of radius r is given by
K.E. of nth orbit
=> (1/k) Ze^{2}/2r
For H atom,
K.E.=(1/4πε) x (e^{2}/2r)
In hydrogen atom the angular momentum of the electron in the lowest energy state is
The angular momentum L =m_{e}vr is on integer multiple of h/2π
mvr= nh/2π
For, n=1
mvr= h/2π
The correct answer is option B.
According to Bohr model of hydrogen atom, the radius of stationary orbit characterized by the principal quantum number n is proportional to
r=(0.529Å)n^{2}/7
r ∝n^{2}
Select an incorrect alternative:
i. the radius of the nth orbit is proprtional to n2
ii. the total energy of the electron in the nth orbit is inversely proportional to n
iii. the angular momentum of the electron in nth orbit is an integral multiple of h/2π
iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy
Statement i. Radius of Bohr's orbit of hydrogen atom is given by
r= n2h2/4π2mKze2
or, r=(0.59A˚)(n2/z)
So, from expression we found r∝n2
Hence the 1st statement is correct.
Statement ii.
We know that
En=13.6 x z2/n2
So, En ∝1/n2
Hence the 2nd statement is wrong.
Statement iii.Bohr defined these stable orbits in his second postulate. According to this postulate:
Hence the 3rd statement is correct.
Statement iv.According to Bohr's theory
Angular momentum of electron in an orbit will be Integral multiple of (h/2π)
Magnitude of potential energy is twice of kinetic energy of electron in an orbit
∣P.E∣=2∣K.E∣
K.E=(13.6ev)( z2/n2)
Hence, The 4th statement is correct.
Bohr used conservation of angular momentum.
For stationary orbits, Angular momentum Iω=nh2π
where n=1,2,3,...etc
In Bohr model of hydrogen atom, radiation is emitted when the electron
The number of times an electron goes around the first Bohr orbit in a second is
We know that,
mvr=h/2π (for first orbit)
⇒mωr^{2}=h2π⇒m×2πv×r^{2}=h/2π
⇒v=h/4π^{2}mr^{2}
The ratio of the volume of the atom and the volume of the nucleus is 10^{15}
The radius of an atomic nucleus is of the order of 10^{−13}cm or 10^{−15}m or one Fermi unit.
On the other hand, the radius of an atom is of the order of 10^{−8}cm or 10^{−10}m or one angstrom unit.
Note:
The radius of nucleus is much smaller than atomic radius.
The ratio of atomic radius to radius of nucleus is 10^{−10}m /10^{−15}m =10^{5}
Volume is proportional to cube of radius.
The ratio of atomic radius to radius of nucleus is (10^{5})^{3}=10^{15}
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