^{th}and (n + 1)

^{th}Bohr’s radius of H-atom is equal to its (n – 1)

^{th}Bohr’s radius. The value of n is:

a) 1

b) 2

c) 3

d) 4

Answer: d

Explanation: r

_{n}∞ n

^{2}

But r

_{n}+ 1 – r

_{n}= r

_{n}– 1

(n + 1)

^{2}– n

^{2}= (n – 1)

^{2}

n = 4

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