The frequency of radiation emitted when theelectron falls from n= 4 to n= 1 in a hydrogen atom will be (Given : ionization energy of H=2.18 × 10^-18 J atom–1 and h= 6.625 × 10^-34 J s)

1.54×10¹⁵s^-1
1.03×10¹⁵s^-1
3.08×10¹⁵s^-1
2.00×10¹⁵s^-1

Solution

\(v=\frac{1}{h}\times IE\times \left [ \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right ]\)

\(=\frac{2.18\times 10^{-18}}{6.625\times 10^{-34}}\times \left [ \frac{1}{1}-\frac{1}{16} \right ]=3.08\times 10^{15}s^{-1}\)

he wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to stationary state 1, would be (Rydberg constant = 1.097×107 m^-1)

406 nm
192 nm
91 nm
9.1×10^-8nm

Solution

\(\frac{1}{\lambda }=R\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )\)

\(\frac{1}{\lambda }=R\left ( \frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right )\)

λ=91.15×10^-9m=91nm

The orbital angular momentum for an electron revolving in anorbit is given by \(\sqrt{\iota (\iota +1)}\cdot \frac{h}{2\pi }\) This momentum for an s-electron will be given by

zero
\(\frac{h}{2\pi }\)
\(\sqrt{2}\frac{h}{2\pi }\)
\(\sqrt{2}\frac{h}{2\pi }\)

Solution

For s-electron, l = 0
∴Orbital angular momentum =\(\sqrt{0(0+1)}\frac{h}{2\pi }=0\)

The energy of second Bohr orbit of the hydrogen atom is -328 kJ mol^-1; hence the energy of fourth Bohrorbit would be:

-41 kJ mol^-1
-82 kJ mol^-1
-164 kJ mol^-1
-1312 kJ mol^-1

Solution

We know that E_n=\(\frac{-1312}{n^{2}}kJ\: mol^{-1}\)
n= 4 (Fourth Bohr orbit)
Given E₄=\(\frac{1312}{4^{2}} –82\: kJ\: mol^{-1}\)

“Electrons are filled in energy orbitals, in increasing order of energy.” This statement is related to

Planck's rule
Hund's rule
Pauli's rule
Aufbau principle

Solution

The given statement is related to Aufbau principle.According to this principle the electrons first occupy the orbital with lowest energy available to them and then enter into higher energy orbitals only when the lower energy orbitals are filled.

The quantum number which is responsible for the size of electron cloud is

spin
azimuthal
principal
magnetic

Solution

Principal quantum number (n) gives the information about the size of electron cloud i.e., the approximate distance of electron cloud from the nucleus

MO configuration of He₂– i

(σ1s)², (σ* 1s)², (σ2s)¹
(σ1s)², (σ* 1s)², (σ* 2s)¹
(σ1s)², (σ* 1s)¹, (σ2s)²
(σ1s)², (σ* 1s)¹, (σ* 2s)²

Solution

He²– contains 5 electrons. So, the molecular orbital electronic configuration is σ1s², σ*1s², σ2s¹

The Bohr orbit radius for the hydrogen atom (n= 1) is approximately 0.530 Å. The radius for the first excited state(n= 2) orbit is (in Å)

0.13
1.06
4.77
2.12

Solution

Given : Radius of hydrogen atom =0.530 Å, Number of excited state (n) = 2 and atomic number of hydrogen atom(Z) = 1. We know that the Bohr radius.

(r)=\(\frac{n^{2}}{Z}\times Radius\: of\: atom=\frac{(2)^{2}}{1}\times 0.530\)

=4×0.530=2.12Å

The radius of hydrogen atom in the ground state is 0.53 Å.The radius of Li²⁺ ion(atomic number = 3)in a similar state is

0.17 Å
0.265 Å
0.53 Å
1.06 Å

Solution

For hydrogen atom (n) = 1(due to ground state)
Radius of hydrogen atom (r) = 0.53 Å.
Atomic number of Li (Z) = 3.

Radius of Li²⁺ion=\(r_{1}\times \frac{n^{2}}{Z}=0.52\times \frac{(1)^{2}}{3}=0.17\)

Which of the following statements do not form a part of Bohr’s model of hydrogen atom ?

Energy of the electrons in the orbits are quantized
The electron(s) in the orbit nearest to the nucleus has thelowest energy
Electrons revolve in different orbitsaround thenucleus
The position and velocity of the electrons in the orbitcannot be determined simultaneously

Solution

This statement is known as uncertainty principle which was given by Heisenberg it is not a Bohr’s postulate.

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