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QUESTION: 1

**Which of the following is correct?**

Solution:

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.

QUESTION: 2

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

Solution:

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.

QUESTION: 3

Wave number of a spectral line for a given transition is x cm^{-1} for He^{+}, then its value for Be^{3+} (isoelectronic of He^{+})for the same transition is

Solution:

QUESTION: 4

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

Solution:

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

QUESTION: 5

An electron in H-atom in its ground state absorbs 1.5 times as much as energy as the minimum required for its escape from the atom

**Q. **

**Thus, kinetic energy given to the emitted electron is**

Solution:

E_{1} = Energy of H-atom in the ground state = 13.6 eV

Energy absorbed = (13.6 x 1.5) = 20.4 eV

E_{2} = Energy of the excited state

= 13.6+ 20.4= 34.0 eV

ΔE = KE = (E_{2} - E_{1})

= 34.0 - 13.6 = 20.4 eV

QUESTION: 6

Ionisation energy of He^{+} is 19.6x10^{-18} J atom ^{-1}. The energy of the first stationary state (n = 1)of Li^{2+} is

Solution:

Ionisation energy = - Energy of the electron

QUESTION: 7

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is (a_{0 }is Bohr radius)

**[AIEEE 2012]**

Solution:

Also, angular momentum is quantised

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 10^{8}ms ^{-1})

**[AIEEE 2012]**

Solution:

For H - atom , Z = 1

QUESTION: 9

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

Solution:

QUESTION: 10

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

Solution:

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

For Lyman series

For shortest wavelength (maximum wave number) n_{2} → ∞

For longest wave length (minimum wave number), n_{2} = (n_{1}, + 1).

For Balmer series, n_{1} = 2

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

Solution:

Angular momentum is quantised , hence

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**

Solution:

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.**

Solution:

Emission (n_{2} = 5 to n_{1} = 2) is called Balmer series and appears in visible region.

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.**

Solution:

QUESTION: 15

lf E_{n} = total energy, Kn = kinetic energy, V_{n} = potential energy and r_{n} = radius of the nth orbit, then based on Bohr’s theory, match the parameter in Column I with the values in Column II.

Solution:

*Answer can only contain numeric values

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?

Solution:

For Bohr radius;

E = -13.6×Z2/ n^{2 }

For H atom, Z=1 and given energy = -3.4

So, we have, -3.4 = -13.6/n^{2}

Or n = 2

*Answer can only contain numeric values

QUESTION: 17

An emission is Be^{3+} in observed at 2.116 A°. In which orbit is it placed?

Solution:

*Answer can only contain numeric values

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?

Solution:

*Answer can only contain numeric values

QUESTION: 19

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

Solution:

For radius r,

circumference of the orbit = 2πr_{n}

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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