ATOMS AND NUCLEI

1 )

A radioactive substance decays to (1 / 16)^th of its initial activity in 40 days. the half life of the radioactive substance expressed in day is

View Solution

Answer :

in t = 40 days, substance decays to (1 / 16) of original i.e. fraction left is

(1 / 16).

If half life is T_(1/2) then fraction left = (N / N₀) = (1 / 2)^{[t / {T(1/2)}]}

∴ (1 / 16) = (1 / 2)^{[(40) / {T(1/2)}]}

∴ (1 / 2)⁴ = (1 / 2)^{[(40) / {T(1/2)}]}

∴ 4 = [(40) / T_(1/2)]

∴ T_(1/2) = (40 / 4) = 10 days.

2 )

If the binding energy of electron in a hydrogen atom is 13.6 eV, the energy required to remove the electron from the second excited state of Li⁺⁺ is

13.6 eV

3.4 eV

30.6 eV

122.4 eV

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Answer :

For atom in any state, E_n = [(13.6) / n²] Z²

For lithium, Z = 3 & for second excited state, n = 3 hence

E_n = [{13.6 × (3)²} / (3)²] = 13.6 eV.

3 )

The radio of minimum to maximum wave length in Balmer series is

(1 / 4)

(5 / 36)

(3 / 4)

(5 / 9)

View Solution

Answer :

For the Balmer series, (1 / λ) = R[(1 / 2²) – (1 / n²)]

for maximum wavelength, (1 / λ_max) = R[(1 / 4) – (1 / 3²)] ----- n = 3

(1 / λ_max) = R[(9 – 4) / (36)]

∴ λ_max = (36 / 5R)

for minimum wavelength n = ∞

∴ (1 / λ_min) = R[(1 / 4) – (1 / ∞)]

(1 / λ_min) = (R / 4)

∴ λ_min = (4 / R)

∴ [(λ_max) / (λ_min)] = (36 / 5R) × (R / 4) = (36 / 20) = (9 / 5)

i.e. [(λ_min) / (λ_max)] = (5 / 9).

4 )

Plutonium decays with half life time 24000 yrs. if Plutonium is stored after 72000 yrs, the fraction of it that remain

(1 / 2)

(1 / 9)

(1 / 12)

(1 / 8)

View Solution

Answer :

given T_(1/2) = 24000 yrs

t = 72000 yrs

we know fraction left = (N / N₀) = (1/2)^{[t / {T(1/2)}]}

hence (N / N₀) = (1 / 2)^{[(72000) / (24000)]} = (1 / 2)³

∴ (N / N₀) = (1 / 8).

5 )

The nuclear ₇N¹⁴ and ₆C¹³ can be described as

Isotones

Isobars

Isotope

Isomer

View Solution

Answer :

if we observe A – Z = (14 – 7) = (13 – 6) = 7

i.e. A – Z is same i.e. number of neutrons are same.

Hence they are isotones.

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