Ncert Solutions Chemistry Class 11th

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New answer posted•a year ago
S
a year ago

The expression for the ionization energy atom,

En = (2.18 * 10-18 x Z2) / n2 J atom-1

For H atom, (Z = 1). So,

En =2.18 * 10-18 * (l)2 J atom-1  (given)

For He+ ion (Z = 2). So,

En =2.18 * 10-18 * (2)2 = 8.72 * 10-18 J atom-1  (one electron specie

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New answer posted•a year ago
S
a year ago

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

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

∴? = 1/ λ = RH Z2 [ (1/22) – (1/42)]

= RH 22 [ (1/22) – (1/42)]

= 3RH /4

For hydrogen spectrum:

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

=> (1/n12) – (1/n22) = 3/4

This corresponds to n1 = 1 and n2 =2 and means

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New answer posted•a year ago
S
a year ago

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 wav

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New answer posted•a year ago
S
a year ago

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

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

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.

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New answer posted•a year ago
S
a year ago

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.

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

This is a Multiple Choice Questions as classified in NCERT Exemplar

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

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New answer posted•a year ago
S
a year ago

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

We know that the amount of work done =-pext? 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 amo

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