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QUESTION: 1

If the electron in H atom jumps from the third orbit to second orbit, the wavelength of the emitted radiation is given by

Solution:

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

QUESTION: 2

The ratio of the speed of the electron in the ground state of hydrogen atom to the speed of light is

Solution:

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

QUESTION: 3

In hydrogen atom the kinetic energy of electron in an orbit of radius r is given by

Solution:

K.E. of nth orbit

=> (1/k) Ze^{2}/2r

For H atom,

K.E.=(1/4πε) x (e^{2}/2r)

QUESTION: 4

In hydrogen atom the angular momentum of the electron in the lowest energy state is

Solution:

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.

QUESTION: 5

According to Bohr model of hydrogen atom, the radius of stationary orbit characterized by the principal quantum number n is proportional to

Solution:

r=(0.529Å)n^{2}/7

r ∝n^{2}

*Multiple options can be correct

QUESTION: 6

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/2

iv. the magnitude of potential energy of the electron in any orbit is greater than its kinetic energy

Solution:

**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

Statement iii.

- An electron revolves around the nucleus in orbits
- The angular momentum of revolution is an integral multiple of h/2π – where Planck’s constant [h = 6.6 x 10-34 J-s].
- Hence, the angular momentum (L) of the orbiting electron is: L = nh/2 π

**Hence the 3rd statement is wrong.
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)

QUESTION: 7

To explain his theory Bohr used:

Solution:

Bohr used conservation of angular momentum.

For stationary orbits, Angular momentum Iω=nh2π

where n=1,2,3,...etc

QUESTION: 8

In Bohr model of hydrogen atom, radiation is emitted when the electron

Solution:

QUESTION: 9

The number of times an electron goes around the first Bohr orbit in a second is

Solution:

We know that,

mvr=h/2π (for first orbit)

⇒mωr^{2}=h2π⇒m×2πv×r^{2}=h/2π

⇒v=h/4π^{2}mr^{2}

QUESTION: 10

The ratio of volume of atom to volume of nucleus is

Solution:

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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