Structure of Atom Test

Result

Structure of Atom Test - 2

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Weekly Quiz Competition

Question 1
4 / -1

Which of the following is correct?

A
₁H¹ and ₂He³ are isotopes

B
₆C¹⁴ and ₇N¹⁴ are isotopes

C
₁₉K³⁹ and ₂₀Ca⁴⁰ are isotones

D
₉F¹⁹ and ₁₁Na²⁴ are isodiaphers

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
4 / -1

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

A
-13 .6 eV atom^-1

B
-27.2 eV atom ^-1

C
+13.6 eV atom ^-1

D
+27.2 eV atom^-1

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
4 / -1

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

A
x cm^-1

B
4x cm^-1

C

D
2x cm^-1

Solution
Question 4
4 / -1

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

A
n = 1 to n = 2

B
n = 2 to n = 4

C
n = 3 to n = 6

D
n = 3 to n = ∞

Solution

Therefore, electronic transition (a) requires greatest quantity of energy.
Question 5
4 / -1

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

A
20.4 eV

B
34.0 eV

C
-20.4 eV

D
-34.0 eV

Solution

E₁ = Energy of H-atom in the ground state = 13.6 eV

Energy absorbed = (13.6 x 1.5) = 20.4 eV
E₂ = Energy of the excited state
= 13.6+ 20.4= 34.0 eV
ΔE = KE = (E₂ - E₁)
= 34.0 - 13.6 = 20.4 eV
Question 6
4 / -1

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

A
4.41 x 10^-16 J atom^-1

B
-4 .41 x 10^-17 J atom^-1

C
-2.20 x 10 ^-15J atom^-1

D
8. 82 x 10^-17J atom^-1

Solution

Ionisation energy = - Energy of the electron
Question 7
4 / -1

The kinetic energy of an electron in the second Bohr orbit of a hydrogen atom is (a₀is Bohr radius)
[AIEEE 2012]

A

B

C

D

Solution

Also, angular momentum is quantised
Question 8
4 / -1

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]

A
1.216x 10^-7 m

B
2.816x 10^-7 m

C
6.500x 10^-7 m

D
8.500x 10^-7 m

Solution

For H - atom , Z = 1
Question 9
4 / -1

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

A
- 54.4 eV

B
-13.6 eV

C
-108.8 eV

D
-27.2 eV

Solution
Question 10
4 / -1

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

A

B

C

D

Solution

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
Question 11
4 / -1

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

A
2πx

B
6πx

C
9x

D
x/3

Solution

Angular momentum is quantised , hence
Question 12
4 / -1

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

A
434.17nm

B
230.20 nm

C
144.70nm

D
289.40 nm

Solution
Question 13
4 / -1

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.

A
UV

B
Visible

C
near IR

D
far IR

Solution

Emission (n₂ = 5 to n₁ = 2) is called Balmer series and appears in visible region.
Question 14
4 / -1

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.

A
a

B
b

C
c

D
d

Solution
Question 15
4 / -1

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.

A
a

B
b

C
c

D
d

Solution
Question 16
4 / -1

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?

A
2

B
3

C
4

D
5

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/n²
Or n = 2
Question 17
4 / -1

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

A
5

B
5

C
8

D
10

Solution
Question 18
4 / -1

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?

A
5

B
7

C
4

D
2

Solution
Question 19
4 / -1

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

A
7

B
11

C
3

D
9

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.