Ncert Solutions Chemistry Class 11th

For an atom? = 1/ λ = R_H Z² [ (1/n₁²) – (1/n₂²)]

For He⁺ spectrum: Z = 4, n₂ = 4, n₁ = 2.

∴? = 1/ λ = R_H Z² [ (1/2²) – (1/4²)]

= R_H 2² [ (1/2²) – (1/4²)]

= 3R_H /4

For hydrogen spectrum:

∴? = 1/ λ = R_H 1² [ (1/n₁²) – (1/n₂²)] = 3R_H /4

=> (1/n₁²) – (1/n₂²) = 3/4

This corresponds to n₁ = 1 and n₂ =2 and means that the transition has taken place in the Lyman series from n = 2 to n =1.

According to Bohr's theory,

mvr = nh / 2π

=> 2πr = nh/mv

=> mv = nh / 2πr - (i)

According to de Broglie equation,

 h / λ - (ii)
Comparing equations (i) and (ii)

nh / 2πr = h / λ

=> 2πr = n λ

Thus, the circumference (2πr) of the Bohr orbit for hydrogen atom is an integral multiple of the de Broglie wavelength.

(a) For n = 4
Total number of electrons = 2n² = 2 * 16 = 32
Half out of these will have m_s = -1/2
∴ Total electrons with m_s (-1/2) = 16
(b) For n = 3
l= 0; m_l = 0, m_s +1/2, -1/2 (two e^–)

(a) The set of quantum numbers is not possible because the minimum value of n can be 1 and not zero.
(b) The set of quantum numbers is possible.
(c) The set of quantum numbers is not possible because, for n = 1, l cannot be equal to 1. It can have 0 value.
(d) The set of quantum numbers is possible.
(e) The set of quantum numbers is not possible because, for n = 3, l cannot be 3.
(f) The set of quantum numbers is possible

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

Specific heat capacity is the quantity of heat required to raise the temperature of one unit mass of a substance by one degree Celsius (or 1 Kelvin). That is why it is an intensive property which does not depend on mass.

This is a Multiple Choice Questions as classified in NCERT Exemplar

option (i)

Variables like p, V, T are called state variables or state functions because their values depend only on the state of the system and not on how it is reached.

This is a Multiple Choice Questions as classified in NCERT Exemplar

Option (iii)

The presence of reactants in a closed vessel made of conducting material e.g., copper or steel is an example of a closed system.

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Thermodynamics is not concerned about how and at what rate these energy transformations are carried out but is based on initial and final states of a system undergoing the change. Laws of thermodynamics apply only when a system is in equilibrium or moves from one equilibrium state to another equilibrium state.

This is a Short Answer Type Questions as classified in NCERT Exemplar

We know that the amount of work done =-p_ext? V

On substituting the values in the formula, we get,

-2bar* (50-10)L=-80Lbar

According to the described problem,1 LBar = 100J

Therefore, -80 L bar= (-80*100)= -8000J

= -8kJ, which is the amount of work done

The significance of the negative sign states that the work is done on the surroundings of the system. In the case of reversible expansion, the work done will be more.

This is a Short Answer Type Questions as classified in NCERT Exemplar

We can conclude from the figure that this change is a reversible change.

Now,

W= -2.303nRT log $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$

But,  p₁V₁ = p₂V_{2 = $\frac{{\mathrm{V}}_2}{{\mathrm{V}}_1}$} = $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$ = $\frac{2}{1}$ =2

 W= -2.303nRT log $\frac{{\mathrm{P}}_1}{{\mathrm{P}}_2}$

 = -2.303 *8.314*1*298*log2

= -2.303 *8.314*298*0.3010J

= -1717.46J