The energy released in the following process is - A - B - C - D -Qmass of A is 1-002 amu - mass of B is 1-004 amu -mass of C is 1-001 amu - mass of D is 1-003 amu

The correct option is D 1.862 MeV Δ m = [1.002 #160;+ 1.004 1.001 1.003] #160; #160; #160; #160; = 0.002 μ E = Q = 0.002× 931.47 ...

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Standard XII

Physics

Mass Energy Equivalence

The energy re...

Question

The energy released in the following process is :
$A + B$ $\to$ $C + D + Q$
(mass of $A$ is $1.002$ amu ; mass of $B$ is $1.004$ amu ;mass of $C$ is $1.001$ amu ; mass of $D$ is $1.003$ amu)

1.254 MeV

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0.931 MeV

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0.465 MeV

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1.862 MeV

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Solution

The correct option is D 1.862 MeV
$Δ m = [1.002 + 1.004 - 1.001 - 1.003]$
$= 0.002 μ$
$E = Q = 0.002 \times 931.478 M e V$
$= 1.862 M e V$

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The energy released in the following process is :A+B →C+D+Q(mass of A is 1.002 amu ; mass of B is 1.004 amu ;mass of C is 1.001 amu ; mass of D is 1.003 amu)

The correct option is D 1.862 MeVΔm=[1.002+1.004−1.001−1.003] =0.002μE=Q=0.002×931.478MeV =1.862MeV

The energy released in the following process is :
$A + B$ $\to$ $C + D + Q$
(mass of $A$ is $1.002$ amu ; mass of $B$ is $1.004$ amu ;mass of $C$ is $1.001$ amu ; mass of $D$ is $1.003$ amu)

The correct option is D 1.862 MeV
$Δ m = [1.002 + 1.004 - 1.001 - 1.003]$
$= 0.002 μ$
$E = Q = 0.002 \times 931.478 M e V$
$= 1.862 M e V$