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Activation Energy

The radioacti...

Question

The radioactivity of a sample is $R_{1}$ at a time $T_{1}$ and $R_{2}$ at a time $T_{2}$ . If the half life of the specimen is T, the number of atoms that have disintegrated in the time $(T_{2} - T_{1})$ is equal to:

(R_{1} T_{1} - R_{2} T_{2})

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(R_{1} - R_{2})

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(R_{1} - R_{2}) / T

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(R_{1} - R_{2}) T / 0.693

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Solution

The correct option is D $(R_{1} - R_{2}) T / 0.693$
The expression for the activity is $A = λ N = \frac{0.693 \times N}{t_{1 / 2}}$
$R_{1} = λ \times N_{1} = \frac{0.693 \times N_{1}}{T}$ ......(1)
$R_{2} = λ \times N_{2} = \frac{0.693 \times N_{2}}{T}$ ......(2)
From equations (1) and (2)
$R_{2} - R_{1} = \frac{0.693 \times N_{2}}{T} - \frac{0.693 \times N_{1}}{T}$
Hence, $(R_{1} - R_{2}) T / 0.693$

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The radioactivity of a sample is R1 at a time T1 and R2 at a time T2. If the half life of the specimen is T, the number of atoms that have disintegrated in the time (T2−T1) is equal to:

The correct option is D (R1−R2)T/0.693The expression for the activity is A=λN=0.693×Nt1/2R1=λ×N1=0.693×N1T......(1)R2=λ×N2=0.693×N2T......(2)From equations (1) and (2)R2−R1=0.693×N2T−0.693×N1THence, (R1−R2)T/0.693

The radioactivity of a sample is $R_{1}$ at a time $T_{1}$ and $R_{2}$ at a time $T_{2}$ . If the half life of the specimen is T, the number of atoms that have disintegrated in the time $(T_{2} - T_{1})$ is equal to: