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Standard XII

Chemistry

Law of Radioactivity

The mass of p...

Question

The mass of proton is 1.0073 $u$ and that of neutron is 1.0087 $u (u =$ atomic mass unit). The ending energy of $_{2} {H e}^{4}$ is (Given, mass of helium necleus $= 4.0015 u)$

28.4

M e V

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62.4

M e V

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42.4

M e V

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10.2

M e V

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Solution

The correct option is A 28.4 $M e V$
$\begin{matrix} Δ m = (2 \times 1.0074 + 2 \times 1.0087 - 4.0015) = 0.0307 E = (Δ m) \times 931 m e V = 0.0307 \times 931 = 28.5 m e V \end{matrix}$
Hence,
option $(A)$ is correct answer.

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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 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 masses of neutron and proton are

1.0087 a . m . u .

and

1.0073 a . m . u .

respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass

4.0015 a . m . u

, then the binding energy fo the helium nucleus will be: (

1 a . m . u . = 931 M e V

)

Q. Find the binding energy of

_{26}^{56} F e

. Atomic mass of

^{56} F e

is 55.9349 u and that of

^{1} H

is 1.00783 u. Mass of neutron = 1.00867 u.

Q. The mass of the proton is 1.0073 u and that of the neutron is 1.0087 u ( u = atomic mass unit). The binding energy of

_{2}^{4} H e

is :

Given, mass of helium nucleus ≈ 4.0015u

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The mass of proton is 1.0073u and that of neutron is 1.0087u(u= atomic mass unit). The ending energy of 2He4 is (Given, mass of helium necleus =4.0015u)

The correct option is A 28.4MeVΔm=(2×1.0074+2×1.0087−4.0015)=0.0307E=(Δm)×931meV=0.0307×931=28.5meVHence,option (A) is correct answer.

The mass of proton is 1.0073 $u$ and that of neutron is 1.0087 $u (u =$ atomic mass unit). The ending energy of $_{2} {H e}^{4}$ is (Given, mass of helium necleus $= 4.0015 u)$

The correct option is A 28.4 $M e V$
$\begin{matrix} Δ m = (2 \times 1.0074 + 2 \times 1.0087 - 4.0015) = 0.0307 E = (Δ m) \times 931 m e V = 0.0307 \times 931 = 28.5 m e V \end{matrix}$
Hence,
option $(A)$ is correct answer.