A hypothetical nuclear fission takes place as following-9236 X -56140 Y -3694 Y - neutronsHow many neutrons will get emitted in the process-A- 1B- 2C- 3D- 4

The correct option is B 2^236_92X → ^140_56Y+^94_36Y+neutrons Z in LHS = 92 Z in RHS = 56 + 36 = 92. A in LHS = 236 A in RHS = 234. A required in RHS = 236 2 ...

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Standard IX

Physics

Nuclear Fission

A hypothetica...

Question

A hypothetical nuclear fission takes place as following:
$_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s$
How many neutrons will get emitted in the process?

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Solution

The correct option is B 2
$_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s$
Z in LHS = 92
Z in RHS = 56 + 36 = 92.

A in LHS = 236
A in RHS = 234.
A required in RHS = 236-234=2.

We know a neutron is $_{0}^{1} n$ .
So, two neutrons will solve the purpose.

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Similar questions

Q. A hypothetical nuclear fission takes place as following:

_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s

How many neutrons will get emitted in the process?

Q. A hypothetical nuclear fission takes place as following:

_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e t r o n / s

How many neutrons will get emitted in the process?

Calculate the number of neutrons emitted when $_{92} U^{235}$ undergoes controlled nuclear fission to $_{54} X e^{142}$ and $_{38} S r^{90} .$

(IIT-JEE, 2010)

Q. The energy of the fast neutrons emitted in a nuclear fission reactor is approximately __________.

Q. The number of neutrons emitted when

_{92}^{235} U

undergoes controlled nuclear fission to

_{54}^{142} Xe

and

_{38}^{90} Sr

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A hypothetical nuclear fission takes place as following: 23692X→14056Y+9436Y+neutrons How many neutrons will get emitted in the process?

The correct option is B 223692X→14056Y+9436Y+neutrons Z in LHS = 92 Z in RHS = 56 + 36 = 92. A in LHS = 236 A in RHS = 234. A required in RHS = 236-234=2. We know a neutron is 10n. So, two neutrons will solve the purpose.

A hypothetical nuclear fission takes place as following:
$_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s$
How many neutrons will get emitted in the process?

The correct option is B 2
$_{92}^{236} X \to_{56}^{140} Y +_{36}^{94} Y + n e u t r o n s$
Z in LHS = 92
Z in RHS = 56 + 36 = 92.

A in LHS = 236
A in RHS = 234.
A required in RHS = 236-234=2.

We know a neutron is $_{0}^{1} n$ .
So, two neutrons will solve the purpose.