A particle carrying 5 electrons starts from rest and is accelerated through a potential difference of 8900 V. Calculate the K.E. acquired by it in MeV. (Charge on electron = 1.6 x 10⁻¹⁹C)
Answers
Answer:
Let us assume particle moved from A to B
Where V
A
- V
B
= 8900V
{V
A
and V
B
is Potential at A and B Respectively}
q = charge on 5 electrons = 5×1.6×10
−19
C
[Also 1eV=1.60×10
−19
J]
Therefore,
Potential energy at A (P.E.
A
) = V
A
×q J
Potential energy at B (P.E.
B
) = V
B
×q J
Kinetic Energy at A (K.E.
A
) = 0 (Since, Particle starts from rest)
Kinetic Energy at B (K.E.
B
) = E
Applying Conservation of Energy,
Total Energy at A = Total Energy at B
K.E.
A
+ P.E.
A
= K.E.
B
+ P.E.
B
→ 0 + V
A
×q = E + V
B
×q
→ E = V
A
×q - V
B
×q
→ E = q × (V
A
- V
B
)
→ E = 5×1.6×10
−19
C × (8900V)
→ E = 71200×10
−19
J
→ E =
1.60×10
−19
71200×10
−19
eV
→ E = 44500 eV
→ E = 0.0445×10
6
eV
→ E = 0.0445 MeV
Explanation:
refer this
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