Answer:

Explanation:
GIVEN
diameter = 15 fm =
m
we use here energy conservation

there will be some initial kinetic energy but after collision kinetic energy will zero

on solving these equations we get kinetic energy initial
J ..............(i)
That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2 e
and gains kinetic energy K =e∆V ..........(ii)
by accelerating through a potential difference ∆V
Thus the alpha particle will
just reach the
nucleus after being accelerated through a potential difference ∆V
equating (i) and second equation we get
e∆V = 35.33 Me V

Answer:
2. [B] = [L]/[T] and [C] = [L]/[T]
Explanation:
I assume you mean this:
A = B² + 2B⁴/C²
Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.
B² = [L]²/[T]²
B = [L]/[T]
B⁴/C² = [L]²/[T]²
C²/B⁴ = [T]²/[L]²
C² = B⁴ [T]²/[L]²
C² = ([L]/[T])⁴ [T]²/[L]²
C² = [L]²/[T]²
C = [L]/[T]
Notice we ignore the 2 coefficient, which is unitless.
Answer:
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