You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard X

Physics

Nuclear Fusion

Consider the ...

Question

Consider the following reaction $^{1} H_{2} +^{1} H_{2} \to_{2} H e^{4} + Q$
If $m (_{1} H^{2}) = 2.0141 u, m (_{2} H e^{4}) = 4.0024 u$ , then the energy Q released (in MeV) in this fusion reaction is

12

No worries! We‘ve got your back. Try BYJU‘S free classes today!

6

No worries! We‘ve got your back. Try BYJU‘S free classes today!

24

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

48

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $24$
$Q =$ $Δ m c^{2}$
$Δ m = 2 m_{H} - m_{H e}$
$= 2 (2.0141) - 4.0024 u$
$= 0.0258 u$
Energy released for $1 a m u$ = 931 MeV
Hence for $Δ m = 0.0258 u$ ,
$E = 0.0258 \times 931 M e V = 24 M e V$

Suggest Corrections

Similar questions

Q. Consider the following reaction

_{1}^{2} H +_{1}^{2} H =_{2}^{4} H e + Q

Mass of the deuterium atom

= 2.0141 u

Mass of helium atom

= 4.0024 u

This is a nuclear ________ reaction in which the energy

Q

released is _______

M e V

Q. Assuming that about 20 MeV of energy is released per fusion reaction

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + E +

other particles, then the mass of

_{1} H^{2}

consumed per day in a fusion reactor of power 1 megawatt will approximately be

Q. The binding energy of deutron

_{1} H^{2}

1.112 M e V

per nucleon and an alpha particle

_{2} H e^{4}

has a binding energy of

7.047 M e V

per nucleon. Then, in the fusion rection

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q

, the energy

Q

released is
(take mass of

_{1} H^{2} = 2.01478 u

;
mass of

_{2} H e^{4} = 4.00338 u

)

Q. The binding energy of deutron

_{1} H^{2}

1.112 M e V

per nucleon and an alpha particle

_{2} H e^{4}

has a binding energy of

7.047 M e V

per nucleon. Then, in the fusion rection

_{1} H^{2} +_{1} H^{2} \to_{2} H e^{4} + Q

, the energy

Q

released is
(take mass of

_{1} H^{2} = 2.01478 u

;
mass of

_{2} H e^{4} = 4.00338 u

)

Q. The amount of energy released in the fusion of

_{1} H^{2}

to form a

_{2} H e^{4}

nucleus will be :
[Binding energy per nucleon of

_{1} H^{2} = 1.1 M e V

and

_{2} H e^{4} = 7 M e V

]

Join BYJU'S Learning Program

Consider the following reaction 1H2+1H2→2He4+QIf m(1H2)=2.0141u,m(2He4)=4.0024u, then the energy Q released (in MeV) in this fusion reaction is

The correct option is C 24Q= Δmc2Δm=2mH−mHe =2(2.0141)−4.0024u =0.0258uEnergy released for 1amu = 931 MeVHence for Δm=0.0258u, E=0.0258×931MeV=24MeV

Consider the following reaction $^{1} H_{2} +^{1} H_{2} \to_{2} H e^{4} + Q$
If $m (_{1} H^{2}) = 2.0141 u, m (_{2} H e^{4}) = 4.0024 u$ , then the energy Q released (in MeV) in this fusion reaction is

The correct option is C $24$
$Q =$ $Δ m c^{2}$
$Δ m = 2 m_{H} - m_{H e}$
$= 2 (2.0141) - 4.0024 u$
$= 0.0258 u$
Energy released for $1 a m u$ = 931 MeV
Hence for $Δ m = 0.0258 u$ ,
$E = 0.0258 \times 931 M e V = 24 M e V$