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Ionic Bond

ntA deuteron ...

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ntA deuteron and an alpha particle are placed in an electric field. The force acting on them are F1 and F2 and their acceleration are a1 and a2 respectively, thenn nt(1) F1 = F2n nt(2) F1 > F2n nt(3) a1 = a2n nt(4) a1 is not equal to a2n

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Q. A deuteron and an

α

-particle are placed in an electric field. The forces acting on them are

F_{1}

F_{2}

, and their accelerations are

a_{1}

a_{2}

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}, a_{3}, . . . . a_{2 n + 1}

are in A.P. then

\frac{a_{2 n + 1} - a_{1}}{a_{2 n + 1} + a_{1}} + \frac{a_{2 n} - a_{2}}{a_{2 n} + a_{2}} + . . . + \frac{a_{n + 2} - a_{n}}{a_{n + 2} + a_{n}}

a_{r} > 0, r ϵ N

a_{1}

a_{2}

a_{3}

a_{2 n}

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

a_{r} > 0, r \in N

a_{1}, a_{2}, a_{3}, . . ., a_{2 n}

are in A.P. then

\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n - 1}}{\sqrt{a_{2}} + \sqrt{a_{3}}} + \frac{a_{3} + a_{2 n - 2}}{\sqrt{a_{3}} + \sqrt{a_{4}}} + . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}

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