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1

AIEEE 2003

MCQ (Single Correct Answer)
The wavelengths involved in the spectrum of deuterium $$\left( {{}_1^2\,D} \right)$$ are slightly different from that of hydrogen spectrum, because
A
the size of the two nuclei are different
B
the nuclear forces are different in the two cases
C
the masses of the two nuclei are different
D
the attraction between the electron and the nucleus is different in the two cases

Explanation

The wavelength of spectrum is given by

$${1 \over \lambda } = R{z^2}\left( {{1 \over {n_1^2}} - {1 \over {n_2^2}}} \right)$$

where $$R = {{1.097 \times {{10}^7}} \over {1 + {m \over M}}}$$

where $$m=$$ mass of electron

$$M=$$ mass of nucleus.

For different $$M,R$$ is different and therefore $$\lambda $$ is different
2

AIEEE 2003

MCQ (Single Correct Answer)
A radioactive sample at any instant has its disintegration rate $$5000$$ disintegrations per minute. After $$5$$ minutes, the rate is $$1250$$ disintegrations per minute. Then, the decay constant (per minute) is
A
$$0.4$$ $$ln2$$
B
$$0.2$$ $$ln2$$
C
$$0.1$$ $$ln2$$
D
$$0.8$$ $$ln2$$

Explanation

$$\lambda = {1 \over t}{\log _e}{{{A_0}} \over A}$$

$$ = {1 \over 5}{\log _e}{{5000} \over {1250}}$$

$$= 0.2{\log _e}4$$

$$ = 0.4{\log _e}2$$
3

AIEEE 2003

MCQ (Single Correct Answer)
A nucleus with $$Z=92$$ emits the following in a sequence: $$$\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,\alpha ,\alpha ,\alpha ,\alpha ,{\beta ^ - },{\beta ^ - },\alpha ,{\beta ^ + },{\beta ^ + },\alpha $$$

Then $$Z$$ of the resulting nucleus is

A
$$76$$
B
$$78$$
C
$$82$$
D
$$74$$

Explanation

The number of $$\alpha $$ - particles released $$=8$$

Therefore the atomic number should decrease by $$16$$

The number of $${\beta ^ - }$$ - particles released $$=4$$

Therefore the atomic number should increase by $$4.$$

Also the number of $${\beta ^ + }$$ particles released is $$2,$$ which

should decrease the atomic number by $$2.$$

Therefore the final atomic number is

$$92-16+4-2=78$$
4

AIEEE 2003

MCQ (Single Correct Answer)
When a $${U^{238}}$$ nucleus originally at rest, decays by emitting an alpha particle having a speed $$'u',$$ the recoil speed of the residual nucleus is
A
$${{4\mu } \over {238}}$$
B
$$ - {{4\mu } \over {234}}$$
C
$$ {{4\mu } \over {234}}$$
D
$$ - {{4\mu } \over {238}}$$

Explanation

Here, conservation of linear momentum can be applied



$$238 \times 0 = 4u + 234v $$

$$\therefore$$ $$v = - {4 \over {234}}u$$

$$\therefore$$ speed $$ = |\overrightarrow v | = {4 \over {234}}u$$

Questions Asked from Atoms and Nuclei

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