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MCQ Practice Test & Solutions: Bohr's Theory Of Hydrogen Atom (19 Questions)

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Test Highlights:

  • - Format: Multiple Choice Questions (MCQ)
  • - Duration: 19 minutes
  • - Number of Questions: 19

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Bohr's Theory Of Hydrogen Atom - Question 1

Which of the following is correct?

Detailed Solution: Question 1

Because isotones means the same number of neutron, So, from the question, option c is right, number of neutron in k is,39-19= 20,and,the number of neutron in ca is also 40-20=20 so it is isotones. 

Bohr's Theory Of Hydrogen Atom - Question 2

‘Hartree’ is the atomic unit of energy and is equal to

Detailed Solution: Question 2

The potential energy of an electron in the first Bohr’s orbit in the H-atom is

This energy is defined as an atomic unit of energy called HARTREE.

Bohr's Theory Of Hydrogen Atom - Question 3

Wave number of a spectral line for a given transition is x cm-1 for He+, then its value for Be3+ (isoelectronic of He+)for the same transition is

Detailed Solution: Question 3



Bohr's Theory Of Hydrogen Atom - Question 4

Which of the following electronic transitions requires that the greatest quantity of energy be absorbed by a hydrogen atom ?

Detailed Solution: Question 4







Therefore, electronic transition (a) requires greatest quantity of energy.

Bohr's Theory Of Hydrogen Atom - Question 5

An electron in a hydrogen atom (ground state) absorbs energy such that the kinetic energy of the emitted electron is 1.5 times the ionization energy (13.6 eV).

Q. Calculate the kinetic energy of the emitted electron.

Detailed Solution: Question 5

Given:

Ionization energy of hydrogen atom (ground state) = 13.6 eV

The question states:
Kinetic energy of emitted electron = 1.5 × ionization energy

Kinetic energy of the emitted electron = 20.4 eV

Bohr's Theory Of Hydrogen Atom - Question 6

Ionisation energy of He+ is 19.6x10-18 J atom -1. The energy of the first stationary state (n = 1)of Li2+ is

Detailed Solution: Question 6

Ionisation energy = - Energy of the electron






Bohr's Theory Of Hydrogen Atom - Question 7

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is (a0 is Bohr radius)

[AIEEE 2012]

Detailed Solution: Question 7


Also, angular momentum is quantised








Bohr's Theory Of Hydrogen Atom - Question 8

Energy of the electron in nth orbit is given by E   Wavelength of light required to excite an electron in an H-atom from level n = 1 to n = 2 will be (h = 6.62 x 10-34 J s ; c = 3.0 x 108ms -1)

[AIEEE 2012]

Detailed Solution: Question 8

For H - atom , Z = 1




Bohr's Theory Of Hydrogen Atom - Question 9

The potential energy of an electron in the second Bohr's orbit of the he±

Detailed Solution: Question 9

Bohr's Theory Of Hydrogen Atom - Question 10

In Lyman series, shortest wavelength of H-atom appears at x m, then longest wavelength in Balmer series of He+ appear at

Detailed Solution: Question 10

For the spectral line in H -atom and H -like species (one electron)

For Lyman series
For shortest wavelength (maximum wave number) n2 → ∞



For longest wave length (minimum wave number), n2 = (n1, + 1).
For Balmer series, n1 = 2


Bohr's Theory Of Hydrogen Atom - Question 11

If the radius of the first Bohr orbit is x, then de-Broglie wavelength of the electron in the third orbit is nearly

Detailed Solution: Question 11

Angular momentum is quantised , hence
 


Bohr's Theory Of Hydrogen Atom - Question 12

Direction (Q. Nos. 12-13) This section contains a paragraph, wach describing  theory, experiments, data etc. three Questions related to paragraph have been given.Each question have only one correct answer among the four given  ptions  (a),(b),(c),(d)

Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Q. 

Wavelength (in nm) associated with this emission is

Detailed Solution: Question 12






Bohr's Theory Of Hydrogen Atom - Question 13

Radius of Bohr’s orbit of H-atom is 52.9 pm. An emission in H-atom starts from the orbit having radius 1.3225 nm and ends at 211.6 pm.

Q.  Spectral line appears in .......... region.

Detailed Solution: Question 13

Emission (n2 = 5 to n1 = 2) is called Balmer series and appears in visible region.

Bohr's Theory Of Hydrogen Atom - Question 14

Direction (Q. Nos. 14 and 15) Choice the correct combination of elements and column I and coloumn II are given as option (a), (b), (c) and (d), out of which ONE option is correct.

Q. 

Match the equation in Column I with the name type in Column II.

Detailed Solution: Question 14

Bohr's Theory Of Hydrogen Atom - Question 15

lf En = total energy, Kn = kinetic energy, Vn = potential energy and rn = radius of the nth orbit, then based on Bohr’s theory, match the parameter in Column I with the values in Column II.

Detailed Solution: Question 15





*Answer can only contain numeric values
Bohr's Theory Of Hydrogen Atom - Question 16

Direction (Q. Nos. 16 - 19) This section contains 4 questions. when worked out will result in an integer from 0 to 9 (both inclusive)

Q. The energy of an electron in the first Bohr orbit of H-atom is -13.6 eV. What is the possible value of quantum number for the excited state to have energy -3.4 eV?


Detailed Solution: Question 16

For Bohr radius;
E = -13.6×Z2/ n
For H atom, Z=1 and given energy = -3.4
So, we have, -3.4 = -13.6/n2
Or n = 2

*Answer can only contain numeric values
Bohr's Theory Of Hydrogen Atom - Question 17

An emission is Be3+ in observed at 2.116 A°. In which orbit is it placed?


Detailed Solution: Question 17


*Answer can only contain numeric values
Bohr's Theory Of Hydrogen Atom - Question 18

At what minimum atomic number, a transition from n = 2 to n = 1 energy level results in the emission of X-rays with wavelength 3.0 x 10-8 m?


Detailed Solution: Question 18






*Answer can only contain numeric values
Bohr's Theory Of Hydrogen Atom - Question 19

Find the number of waves made by a Bohr’s electron in one complete revolution in its 3rd orbit


Detailed Solution: Question 19

For radius r,
circumference of the orbit = 2πrn


Number of waves in one complete revolution = nλ
For third orbit = nλ = 3λ
Thus, three waves are formed in one revolution. 

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