Fusion processes, like combining two deuterons to form a `He` nucleus are impossible at ordinary temperature and pressure. The reasons for this can be traced to the fact:
Fusion processes, like combining two deuterons to form a `He` nucleus are impossible at ordinary temperature and pressure. The reasons for this can be traced to the fact:
A
nuclear forces have short range
B
nuclei are positively charged
C
the original nuclei must be completely ionized before fusion can take place
D
the original nuclei must first break up before combining with each other
Text Solution
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The correct Answer is:
A, B
Fusion processes are impossible at ordinary temperature and pressures. This is because nuclei are positively charged and nuclear forces are the strongest forces having short range. Choice (a) and (b) is correct.
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Knowledge Check
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Calculate the half-life period of a radioactive element which remains only 1//16 of its original amount in 4740 years:
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Calculate the half-life period of a radioactive element which remains only 1//16 of its original amount in 4740 years:
A
1185 years
B
2370 years
C
`52.5` years
D
none of these
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) What is the activity in Ci (curie) of 1.0mole plutonium -239 ? (t_(1//2)=24000 yeasrs)
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) What is the activity in Ci (curie) of 1.0mole plutonium -239 ? (t_(1//2)=24000 yeasrs)
A
`1.49` Ci
B
`14.9` Ci
C
`5.513xx10^(11)` Ci
D
`None of these
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Half-life period of U^(2.5xx10^(5) years. In how much thime will the amount of U^(237) remaining be only 25% of the original amount ?
Radioactive disintegration is a first order reaction and its rate depends only upon the nature of nucleus and does not depend upon external factors like temperature and pressure. The rate of radioactive disintegration (Activity) is represented as -(dN)/(dt)=lambdaN Where lambda= decay constant, N= number of nuclei at time t, N_(0) =intial no. of nuclei. The above equation after integration can be represented as lambda=(2.303)/(t)log((N_(0))/(N)) Half-life period of U^(2.5xx10^(5) years. In how much thime will the amount of U^(237) remaining be only 25% of the original amount ?
A
`2.5xx10^(5)` year
B
`1.25xx10^(5)` years
C
`5xx10^(5)` years
D
none of these
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