The minimum kinetic energy needed by an alpha particle to cause the nuclear reaction _7^16 ~N+ _2^4 He arrow _1^1 H+ _8^19 O in a laboratory frame is n (in M e V . Assume that _7^16 ~N is at rest in the laboratory frame. The masses of _7^16 ~N, _2^4 He, _1^1 H and _8^19 O can be taken to be 16.006 u, 4.003 u, 1.008 u and 19.003 u , respectively, where 1 u=930 \,MeVc^-2 . The value of n is \underline\hspace2cm .

Atoms and Nuclei PYQ — JEE Advanced 2022 · Shift 1 · Q38 with Solution

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Correct answer: 2.30TO2.35

Identify the reaction and compute the Q-value

The reaction is

{}^{16}_{7}\mathrm N + {}^{4}_{2}\mathrm{He} \rightarrow {}^{1}_{1}\mathrm H + {}^{19}_{8}\mathrm O

with ${}^{16}_{7}\mathrm N$ initially at rest in the laboratory frame.

For a nuclear reaction,

Q = \left(m_{\text{initial}}-m_{\text{final}}\right)c^2

Given masses:

$m\left({}^{16}_{7}\mathrm N\right)=16.006\,u$
$m\left({}^{4}_{2}\mathrm{He}\right)=4.003\,u$
$m\left({}^{1}_{1}\mathrm H\right)=1.008\,u$
$m\left({}^{19}_{8}\mathrm O\right)=19.003\,u$

Initial mass:

M_i = 16.006+4.003=20.009\,u

Final mass:

M_f = 1.008+19.003=20.011\,u

Hence,

\Delta m = M_i-M_f = 20.009-20.011=-0.002\,u

So,

Q = (-0.002)(930)\,\text{MeV} = -1.86\,\text{MeV}

Since $Q<0$ , this is an endothermic reaction, so a threshold kinetic energy is required.

Threshold energy formula in laboratory frame

For a reaction

a + A \rightarrow b + B

with target $A$ at rest, the minimum projectile kinetic energy is

K_{\min} = -Q\left(1+\frac{m_a}{m_A}\right)

for a two-body final state.

Here,

projectile $a = {}^{4}_{2}\mathrm{He}$ with mass $m_a=4.003\,u$
target $A = {}^{16}_{7}\mathrm N$ with mass $m_A=16.006\,u$

Thus,

K_{\min} = 1.86\left(1+\frac{4.003}{16.006}\right)

Now,

\frac{4.003}{16.006} \approx 0.2501

So,

K_{\min} \approx 1.86(1.2501)

K_{\min} \approx 2.325\,\text{MeV}

Final answer

Therefore,

n \approx 2.33

MeV.

As an integer-type/numerical entry with the given stored range, the value is approximately:

\boxed{2.33\,\text{MeV}}

More from Atoms and Nuclei

View written solutionFree

Correct answer: 2.30TO2.35

Identify the reaction and compute the Q-value

The reaction is

{}^{16}_{7}\mathrm N + {}^{4}_{2}\mathrm{He} \rightarrow {}^{1}_{1}\mathrm H + {}^{19}_{8}\mathrm O

with ${}^{16}_{7}\mathrm N$ initially at rest in the laboratory frame.

For a nuclear reaction,

Q = \left(m_{\text{initial}}-m_{\text{final}}\right)c^2

Given masses:

$m\left({}^{16}_{7}\mathrm N\right)=16.006\,u$
$m\left({}^{4}_{2}\mathrm{He}\right)=4.003\,u$
$m\left({}^{1}_{1}\mathrm H\right)=1.008\,u$
$m\left({}^{19}_{8}\mathrm O\right)=19.003\,u$

Initial mass:

M_i = 16.006+4.003=20.009\,u

Final mass:

M_f = 1.008+19.003=20.011\,u

Hence,

\Delta m = M_i-M_f = 20.009-20.011=-0.002\,u

So,

Q = (-0.002)(930)\,\text{MeV} = -1.86\,\text{MeV}

Since $Q<0$ , this is an endothermic reaction, so a threshold kinetic energy is required.

Threshold energy formula in laboratory frame

For a reaction

a + A \rightarrow b + B

with target $A$ at rest, the minimum projectile kinetic energy is

K_{\min} = -Q\left(1+\frac{m_a}{m_A}\right)

for a two-body final state.

Here,

projectile $a = {}^{4}_{2}\mathrm{He}$ with mass $m_a=4.003\,u$
target $A = {}^{16}_{7}\mathrm N$ with mass $m_A=16.006\,u$

Thus,

K_{\min} = 1.86\left(1+\frac{4.003}{16.006}\right)

Now,

\frac{4.003}{16.006} \approx 0.2501

So,

K_{\min} \approx 1.86(1.2501)

K_{\min} \approx 2.325\,\text{MeV}

Final answer

Therefore,

n \approx 2.33

MeV.

As an integer-type/numerical entry with the given stored range, the value is approximately:

\boxed{2.33\,\text{MeV}}

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