Practice Start Test

Atoms and Nuclei

NEET Physics · 119 questions · Page 12 of 12 · Click an option or "Show Solution" to reveal answer

Q111
Half life of a radioactive element is 12.5 hour and its quantity is 256 g. After how much time its quantity will remain 1 g?
A 50 hrs
B 100 hrs
C 150 hrs
D 200 hrs
Correct Answer
Option B
Solution

N = N 0

(12)n{\left( {{1 \over 2}} \right)^n}

⇒\Rightarrow 1 = 256

(12)n{\left( {{1 \over 2}} \right)^n}

⇒\Rightarrow n = 8 Time for 8 half life = 100 hours

Q112
X(n, α\alpha ) 37{}_3^7Li, then X will be
A 510{}_5^{10} B
B 59{}_5^9 B
C 511{}_5^{11} Be
D 24{}_2^4 He.
Correct Answer
Option A
Solution
510{}_5^{10}

B +

01{}_0^{1}

n →\to

24{}_2^{4}

He +

37{}_3^{7}

Li

Q113
Which rays contain (positive) charged particles ?
A α\alpha -rays
B β\beta -rays
C γ\gamma -rays
D X-rays
Correct Answer
Option A
Solution

α\alpha-rays contain Helium nuclei which contains 2 unit of positive charge.

Q114
The energy of hydrogen atom in n th orbit is E n then the energy in n th orbit of singly ionised helium atom will be
A 4E n
B E n /4
C 2E n
D E n /2
Correct Answer
Option A
Solution

As E n =

−2π2mK2Z2e4n2h2{{ - 2{\pi ^2}m{K^2}{Z^2}{e^4}} \over {{n^2}{h^2}}}

For helium Z = 2. Hence requisite answer is 4E n .

Q115
Maximum frequency of emission is obtained for the transition
A n = 2 to n = 1
B n = 6 to n = 2
C n = 1 to n = 2
D n = 2 to n = 6.
Correct Answer
Option A
Solution

ν\nu ∝\propto

[1n12−1n22]\left[ {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right]

Maximum frequency of emission happens when n 1 = 1 and n 2 = 2

Q116
The life span of atomic hydrogen is
A fraction of one second
B one year
C one hour
D one day
Correct Answer
Option A
Solution

Atomic hydrogen is unstable and it has life period of a fraction of a second.

Q117
The relation between λ\lambda and T 1/2 as (T 1/2 → \to half life)
A T 1/2 = ln⁡2λ{{\ln 2} \over \lambda }
B T 1/2 ln2 = λ\lambda
C T 1/2 = 1λ{1 \over \lambda }
D (λ\lambda + T 1/2 ) = ln2
Correct Answer
Option A
Solution

The relation is : T 1/2 =

ln⁡2λ{{\ln 2} \over \lambda }
Q118
When an electron does transition from n = 4 to n = 2, then emitted line spectrum will be
A first line of Lyman series
B second line of Balmer series
C first line of Balmer series
D second line of Paschen series.
Correct Answer
Option B
Solution

Jump to second orbit leads to Balmer series. The jump from 4th orbit shall give rise to second line of Balmer series.

Q119
Nuclear fission is best explained by
A liquid droplet theory
B Yukawa π\pi -meson theory
C independent particle model of the nucleus
D proton-proton cycle.
Correct Answer
Option A
Solution

According to liquid drop model of nucleus, an excited nucleus breaks into lighter nuclei just like an excited drop breaks into tiny drops.

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