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Potential Energy - External Viewpoint

A deuteron an...

Question

A deuteron and an $α$ -particle are placed in an electric field. The forces acting on them are $F_{1}$ and $F_{2}$ , and their accelerations are $a_{1}$ and $a_{2}$ respectively.

A

F_{1} = F_{2}

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B

F_{1} \neq F_{2}

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C

a_{1} = a_{2}

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D

a_{1} \neq a_{2}

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Solution

The correct options are
B $F_{1} \neq F_{2}$
C $a_{1} = a_{2}$

Deuteron(charge+e mass 2m) and alpha particle(charge+2e, mass 4m)Since the charge of deuteron and $α$ -particle are different, the forces acting are different.

Thus

$\frac{F_{1}}{F_{2}} = \frac{q_{1}}{q_{2}} = \frac{1}{2}$

Also as their masses are different we get

$\frac{F_{1}}{F_{2}} = \frac{m_{1} a_{1}}{m_{2} a_{2}} = \frac{1}{2}$

or

$\frac{a_{1}}{a_{2}} = 1$

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Similar questions

Q. Assertion :A deuteron and an

α -

particle are placed in an electric field. If

F_{1}

F_{2}

be the forces acting on them and

a_{1}

a_{2}

be their accelerations respectively then,

a_{1} = a_{2}

. Reason: Forces will be same in electric field.

a_{1}, a_{2}, . . . .

are in H.P. and

f (k) = n \sum r = 1 (a_{1} = a_{i})

\frac{a_{1}}{f (1)}, \frac{a_{2}}{f (2)}, . . . \frac{a_{n}}{f (n)}

Q. In the figure, the blocks

A, B

C

m

each have acceleration

a_{1}, a_{2}

a_{3}

F_{1}

F_{2}

are external forces of magnitudes

2 m g

each. What is the relation between the acceleration of the masses

A

B

C

a_{1}, a_{2}, a_{3}

f_{k} = \sum_{r = 1}^{n} a_{r} - a_{k}

2^{α_{1}}, 2^{α_{2}}, 2^{α_{3}}, 2^{α_{4}}, . . . .

α_{1} = \frac{a_{1}}{f_{1}}, α_{2} = \frac{a_{2}}{f_{2}}, α_{3} = \frac{a_{3}}{f_{3}} . . . .

}

Q. In the figure, the block

A, B

C

m

have accelerations

a_{1}, a_{2}

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F_{2}

are external forces of magnitude

2 m g

m g

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