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The half life period of a radioactive element X is same as the mean life time of another radioactive element Y. Initially both of them have the same number of atoms. Then

X and Y have the same decay rate initially

X and Y decay at the same rate always

Y will decay at a faster rate than X

X will decay at a faster rate than Y

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Explanation

$\Rightarrow \frac{0.693}{λ_{x}} = \frac{1}{λ_{y}} \Rightarrow λ_{x} = 0.693 λ_{y} or λ_{x} < λ_{y}$

Also rate of decay = $λ$ N
Initially number of atoms (N) of both are equal but since $λ_{y} > λ_{x}$ therefore, y will decay at a faster rate than x.

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After 280 days, the activity of a radioactive sample is 6000 dps. The activity reduces to 3000 dps after another 140 days. The initial activity of the sample in dps is

6000

9000

3000

24000

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Explanation

(d) Here the activity of the radioactive sample reduces to half in 140 days. Therefore, the half life of the sample is 140 days. 280 days is it’s two half lives. So before two half lives it’s activity was ( $2^{2} \times$ present activity).
$∴$ Initial activity = $2^{2} \times$ 6000=24000 dps

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Excitation energy of a hydrogen like ion in its first excitation state is 40.8 eV. Energy needed to remove the electron from the ion in ground state is

54.4 eV

13.6 eV

40.8 eV

27.2 eV

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Explanation

(a) Excitation energy

$∆ E = E_{2} - E_{1} = 13.6 Z^{2} [\frac{1}{1^{2}} - \frac{1}{2^{2}}]$

$\Rightarrow 40.8 = 13.6 \times \frac{3}{4} \times Z^{2} \Rightarrow Z = 2$

Now required energy to remove the electron from ground state

$= \frac{+ 13.6 Z^{2}}{1^{2}} = 13.6 Z^{2} = 54.4 eV$

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Consider a hydrogen like atom whose energy in nth exicited state is given by $E_{n} = - \frac{13.6 Z^{2}}{n^{2}}$ when this excited atom makes a transition from excited state to ground state, most energetic photons have energy $E_{\max}$ = 52.224 eV and least energetic photons have energy $E_{\min}$ = 1.224 eV. The atomic number of atom is

None of these

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Explanation

(a) Maximum energy is liberated for transition $E_{n} \to E_{1}$ and minimum energy for $E_{n} \to E_{n - 1}$
Hence $\frac{E_{1}}{n^{2}} - E_{1} = 52.224 eV$ ……(i)
and $\frac{E_{1}}{n^{2}} - \frac{E_{1}}{{(n - 1)}^{2}} = 1.224 eV$ …..(ii)
Solving equations (i) and (ii) we get
and $E_{1} = - 54.4 eV$ and n = 5
Now $E_{1} = - \frac{13.6 Z^{2}}{1^{2}} = - 54.4 eV$ . Hence Z = 2

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A radioactive sample is $α$ -emitter with half life 138.6 days is observed by a student to have 2000 disintegration/sec. The number of radioactive nuclei for given activity are

$3.45 \times 10^{10}$

$1 \times 10^{10}$

$3.45 \times 10^{15}$

$2.75 \times 10^{11}$

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Explanation

(a) Activity of substance that has 2000 disintegration/sec

The number of radioactive nuclei having activity A
$N = \frac{A}{λ} = \frac{2000 \times T_{1 / 2}}{\log_{e} 2} = \frac{2000 \times 138.6 \times 24 \times 3600}{0.693} = 3.45 \times 10^{10}$

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The ratio of radii of nuclei ${}_{13}^{27}{Al}$ and ${}_{52}^{A}X$ is 3 : 5. The number of neutrons in the nuclei of X will be

125

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Explanation

(b) $r \propto A^{1 / 3} \Rightarrow \frac{r_{1}}{r_{2}} = {(\frac{A_{1}}{A_{2}})}^{1 / 3}$

$\Rightarrow \frac{3}{5} = {(\frac{27}{A})}^{1 / 3} \Rightarrow \frac{27}{125} = \frac{27}{A} \Rightarrow A = 125$

Number of nuclei in atom X = A-52 = 125-52 = 73

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In a common emitter transistor amplifier, the audio signal voltage across resistance of 1 k $Ω$ is 2V. If the base resistance is 200 $Ω$ and the current amplification factor is 50, the input signal voltage will be

4 mV

8 mV

16 mV

32 mV

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Explanation

$I_{c} = \frac{V_{c}}{R_{c}} = \frac{2}{1000} = 2 m A C u r r e n t a m p l i f i c a t i o n f a c t o r, β = \frac{I_{c}}{I_{B}} \Rightarrow I_{B} = \frac{I_{c}}{β} = 4 \times 10^{- 2} m A V_{B} = I_{B} \times R_{B} = 4 \times 10^{- 2} \times 200 = 8 m V$

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If a small amount of aluminium is added to the silicon crystal

Its resistance is decreased

It becomes a p-type semiconductor

There will be less free-electron than holes in semiconductor

All of these are correct

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Explanation

If trivalent impurity added to the pure semiconductor, then a p-type semiconductor is formed. In a p-type semiconductor, the majority charge carriers are holes and the minority charge carriers are electrons.

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A transistor is operated in common emitter configuration at $V_{c} = 10 V .$ When base current is changed from 10mA to 30mA, it produces a change in emitter current from 2A to 4A, the current amplification factor is

100

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Explanation

$(2) β = \frac{∆ I_{c}}{∆ I_{B}} = \frac{∆ I_{E} - ∆ I_{B}}{∆ I_{B}} = \frac{∆ I_{E}}{∆ I_{B}} - 1 = \frac{2}{20 x 10^{- 3}} - 1 = \frac{200}{2} - 1 = 99$

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When a p-n junction is forward biased:

Depletion region becomes thick

p-side is at a higher potential than n side

Current flowing is zero

Effective resistance is of the order of $10^{6} W$

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Explanation

In F.B., p-region is at higher potential and n-region is to be at the lower potential.

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