Nuclear Chemistry — NEET UG practice

8 questions

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Sample questions with solutions

Q1 · 2022

The half life of a first order reaction is 2000 years. If the concentration after 8000 years is 0.02 M, then the initial concentration was :

A.
0.04 M
B.
0.16 M
C.
0.32 M
D.
0.08 M

Answer: C

For a first-order reaction, the concentration is reduced to half of its previous value after each half-life period, regardless of the starting concentration.
Finding the number of half-lives that have passed in the total time given:

n = \frac{\text{Total time}}{\text{Half-life}} = \frac{8000}{2000} = 4

Since the concentration halves with each half-life, after $n$ half-lives, the remaining concentration relates to the initial concentration $[A]_0$ by:

[A]_t = \frac{[A]_0}{2^n}

Given that the concentration after 4 half-lives is $0.02\ M$ , we substitute the known values to solve for the initial concentration.

0.02 = \frac{[A]_0}{2^4}

Therefore, solving for $[A]_0$ :

[A]_0 = 0.02 \times 2^4 = 0.02 \times 16 = 0.32\ M

Hence, the initial concentration was 0.32 M, so the answer is option C.

Q2 · 2005

A nuclide of an alkaline earth metal undergoes radioactive decay by emission of the periodic table to which the resulting daughter element would belong is

A.
Gr.13
B.
Gr. 17
C.
Gr. 14
D.
Gr. 16

Answer: C

When a radioactive nucleus emits an alpha particle ( $\alpha = {}_2^4He$ ), it loses 2 protons and 2 neutrons, so the atomic number decreases by 2 and the mass number decreases by 4.
Since the atomic number decreases by 2, the resulting element shifts two groups to the left in the periodic table (in the older group numbering, this means shifting by two units).
Starting from an alkaline earth metal in Group 2 (IIA), emitting one alpha particle shifts the daughter element by 2 groups, effectively into what corresponds to Group 16 territory in the old numbering.
However, since this is part of a radioactive decay series (as noted, involving successive emissions), continued alpha and beta emissions eventually shift the atomic number further until it stabilizes.
The known stable end product of such natural radioactive decay series (like those starting from radium) is lead (Pb), which belongs to Group 14.
Hence, the resulting daughter element belongs to Group 14, so the answer is option C.

Q3 · 2004

The radioactive isotope $_{27}^{60}Co$ which is used in the treatment of cancer can be made by (n,p) reaction. For this reaction the target nucleus is

A.
$_{28}^{59}Ni$
B.
$_{27}^{59}Co$
C.
$_{28}^{60}Ni$
D.
$_{27}^{60}Co$

Answer: C

In an (n, p) reaction, a target nucleus absorbs a neutron and, in response, releases a proton, changing its identity in the process.
Since a proton is ejected, the atomic number of the target decreases by 1 (one fewer proton) while the mass number stays the same (since a neutron replaces the lost proton in mass terms).

{}_Z^AX + {}_0^1n \rightarrow {}_{Z-1}^{A}Y + {}_1^1p

Applying this to the known product, ${}_{27}^{60}Co$ , we work backward: since the product has atomic number 27, the target nucleus must have had atomic number 28 (one more, since it lost a proton to become 27), while keeping the same mass number, 60.

{}_{28}^{60}Ni + {}_0^1n \rightarrow {}_{27}^{60}Co + {}_1^1p

Therefore, the target nucleus must be ${}_{28}^{60}Ni$ .
Hence, the answer is option C.

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