Bombardment of lithium with proton gives rise to the following reaction - 73Li-11 H -2 42He - QThe Q-value is atomic masses of lithium- proton and helium are 7-016 amu- 1-008 amu and 4-004 amu respectively

The correct option is C 14.904 MeV ^7_3Li + #160; #160; ^1_1H → #160;2 #160; #160; ^4_2He + Q Mass defect = (7.016 + 1.008 2× 4.004) ...

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Standard XII

Physics

Binding Energy

Bombardment o...

Question

Bombardment of lithium with proton gives rise to the following reaction :
$_{3}^{7} L i +_{1}^{1} H \to$ 2( $_{2}^{4} H e) + Q$
The Q-value is
(atomic masses of lithium, proton and helium are $7.016$ amu, $1.008$ amu and $4.004$ amu respectively)

14.904

MeV

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16.774

MeV

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10.634

MeV

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18.633

MeV

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Solution

The correct option is C $14.904$ MeV
$_{3}^{7} L i +_{1}^{1} H \to 2_{2}^{4} H e + Q$
Mass defect $= (7.016 + 1.008 - 2 \times 4.004) μ$ $= 0.016 a m u$

$1 a m u = 931.478 \frac{M e V}{C^{2}}$
$E = 0.016 \times 931.478 \frac{M e V}{C^{2}} \times C^{2}$ $= 14.903 M e V$

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Similar questions

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. The binding energy per nucleon of

_{3}^{7} Li

and

_{2}^{4} He

nuclei are

5.60 MeV

and

7.06 MeV

,respectively. In the nuclear reaction

_{3}^{7} Li +_{1}^{1} H \to_{2}^{4} He + Q,

the value of energy

Q

released is

Q. The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu. The binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. Find Binding energy of an

α -

particle in

M e V

[m_{p r o t o n} = 1.007825 a m u, m_{n e u t r o n} = 1.008665 a m u, m_{h e l i u m} = 4.002800 a m u]

Q. Find the binding energy of an

α -

particle from the following data.
Mass of the helium nucleus

= 4.001265 a m u

Mass of proton

= 1.007277 a m u

Mass of neutron

= 1.00866 a m u

(take

1 a m u = 931.4813 M e V)

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Bombardment of lithium with proton gives rise to the following reaction :73Li+11H→2(42He)+QThe Q-value is (atomic masses of lithium, proton and helium are 7.016 amu, 1.008 amu and 4.004 amu respectively)

The correct option is C 14.904 MeV73Li+11H→242He+QMass defect =(7.016+1.008−2×4.004)μ=0.016amu1amu=931.478MeVC2E=0.016×931.478MeVC2×C2=14.903MeV

Bombardment of lithium with proton gives rise to the following reaction :
$_{3}^{7} L i +_{1}^{1} H \to$ 2( $_{2}^{4} H e) + Q$
The Q-value is
(atomic masses of lithium, proton and helium are $7.016$ amu, $1.008$ amu and $4.004$ amu respectively)

The correct option is C $14.904$ MeV
$_{3}^{7} L i +_{1}^{1} H \to 2_{2}^{4} H e + Q$
Mass defect $= (7.016 + 1.008 - 2 \times 4.004) μ$ $= 0.016 a m u$

$1 a m u = 931.478 \frac{M e V}{C^{2}}$
$E = 0.016 \times 931.478 \frac{M e V}{C^{2}} \times C^{2}$ $= 14.903 M e V$