Solution

The energy of an electron in the first Bohr orbit of H atom is –13.6 eV. The possible energy value(s) of the excited state(s)for electrons in Bohr orbits of hydrogen is (are)

–3.4 eV
–4.2 eV
–6.8 eV
Both (a) and (c)

Solution

The energy of an electron on Bohr orbits of hydrogen atoms is given by the expression

\(E_{n} = \frac{-constant}{n^{2}}\)

Where n takes only integral values. For the first Bohr orbit, n = 1 and it is given that E₁ = –13.6 eV

Hence \(E_{n} = \frac{-13.6 \; ev}{n^{2}}\)

Now out of given values of energy,only –3.4 eV can be obtained by substituting n = 2 in the above expression.

The number of radial nodes of 3s and 2p orbitals are respectively

2, 0
0, 2
1, 2
2, 1

Solution

Number of radial nodes = (n - l - 1)

For 3s: n = 3, l = 0 (Number of radial node = 2)

For 2p: n = 2, l = 1 (Number of radial node = 0)

Uncertainty in the position of an electron (mass = 9.1 × 10^–31kg)moving with a velocity 300 ms^-1, accurate upto 0.001% will be(h = 6.63 × 10^-34Js)

1.92 × 10^-2 m
3.84 × 10^-2 m
19.2 × 10^-2 m
5.76 × 10^-2 m

Solution

Given m = 9.1 × 10^-31 kg

h = 6.6 × 10^-34 Js

\(\Delta v = \frac{300 \times 0.001}{100} = 0.003 \; ms^{-1}\)

From Heisenberg's uncertainity principle

\(\Delta x = \frac{6.62 \times 10^{-34}}{4 \times 3.14 \times 0.003 \times 9.1 \times 10^{-31}} = 1.92 \times 10^{-2} \; m\)

Which two orbitals are both located between the axes of coordinate system, and not along the axes?

d_xy, d_z²
d_yz, p_x
d_{x² - y²}, p_z
none of these

Li and a proton are accelerated by the same potential, their de Broglie wavelengths λ_Li and λp have the ratio (assume m_Li = 9m_p)

1 : 2
1 : 4
1 : 1
1 : 3√3

Solution

In a multi-electron atom, which of the following orbitals described by the three quantum members will have the same energy in the absence of magnetic and electric fields?

(A) n = 1, l = 0, m = 0

(B) n = 2, l = 0, m = 0

(D) n = 3, l = 2, m = 1

(E) n = 3, l = 2, m = 0

(D) and (E)
(C) and (D)
(B) and (C)
(A) and (B)

Solution

The energy of an orbital is given by (n + l) in (D) and (E).

(n + l) value is (3 + 2) = 5 hence they will have same energy, since there n values are also same.(h = 6.63 × 10^–34Js)

Consider the ground state of Cr atom (Z = 24). The number of electrons with the azimuthal quantum numbers, l = 1 and 2 are, respectively

16 and 4
12 and 5
12 and 4
16 and 5

Solution

Electronic configuration of Cr atom (Z = 24)

= 1s², 2s² 2p⁶, 3s² 3p⁶ 3d⁵, 4s¹

when l = 1, p-subshell,

Numbers of electrons = 12

when l = 2, d-subshell,

Numbers of electrons = 5

In Bohr series of lines of hydrogen spectrum, the third line from the red end corresponds to which one of the following inter-orbit jumps of the electron for Bohr orbits in an atom of hydrogen

5 → 2
4 → 1
2 → 5
3 → 2

Solution

The lines falling in the visible region comprise Balmer series. Hence the third line from red would be n₁ = 2, n₂ = 5 i.e, 5 → 2.

Kinetic energy of an electron in hydrogen atom increases after transition from an orbit n₁ to another orbit n₂. Then

n₁ < n₂
n₁ > n₂
it is not possible to predict
none of these

Solution

K.E. of an electron in a Bohr orbit is equal to the magnitude of the total energy but of opposite sign. So it varies inversly to the square of principal quantum number.

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