IIT JEE , AIEEE Forum - Chemistry , Physical Chemistry

mastermind890

the difference between the nth and(n+1)th Bohr's radius of H atom is equal to its(n-1)th Bohr's radius .find the value of n??

mastermind890

plz reply.....

sandeepramesh

can you tell the formula for the radius in the nth orbit? I dont seem to remember it! after that itll be straightforward

tarun_bits

n - n-1=n-1
-1=n-1
n=0

i dont know what hav i done

somyekathait

is it 4

punnima

r_n = e₀h²n²/ pmZe²
r_n+1- r_n = e₀h²(n+1)²   - e₀h²n²      ===> LHS
                 pmZe²          pmZe²
r_n-1= e₀h²(n-1)²                   ===> RHS
             pmZe²equating LHS and RHS
-n² +(n+1)² =(n-1)²n²-n² +1+2n=n²+1-2n

n²-4n=0

=>n=0 (or) n=4
Ans: n=4

punnima

if we take r_n- r_n+1 = r_n-1

we get n²+2=0
or n=+

2

n=1.414

1

hence 2 ans r possible
either n=4 or n=1

punnima

is this d ans??
if yes, plizz rate

raulrag009

Here is my solution

$As\;r_{n}=(.529*n^{2}) ang.\\\\ r_{n+1}-r_{n}=r_{n-1}\\\\ (n+1)^{2}-n^{2}=(n-1)^{2}\\\\ n=0\quad\;n=4$

somyekathait

i agree with raulrag007 's answer

mastermind890

even i have got 4 but the answer given is 2.

somyekathait

but even if u substitute value of n as 2 in the question itself the answer is not coming to be right

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