When a transition of electron in H e+takes place from n2 to n1 then wave number in terms of Rydberg constant R will be ( Given n1+n2=4, n2-n1=2 )

Question

When a transition of electron in H e+takes place from n2 to n1 then wave number in terms of Rydberg constant R will be ( Given n1+n2=4, n2-n1=2 ) (A) (3 R/4) (B) (8 R/9) (C) (32 R/9) (D) (24 R/9)

Tardigrade Admin · Accepted Answer

n1+n2=4 n2-n1=2 2n2=6 n2=3 n1=1 3 arrow 1 =R Z2((1/n12)-(1/n22)) =R × 22((1/12)-(1/32)) =4 R × (8/9) =(32 R/9)

Q. When a transition of electron in $H e^{+}$ takes place from $n_{2}$ to $n_{1}$ then wave number in terms of Rydberg constant $R$ will be
( Given $n_{1} + n_{2} = 4, n_{2} - n_{1} = 2$ )

$\frac{3 R}{4}$

$\frac{8 R}{9}$

$\frac{32 R}{9}$

$\frac{24 R}{9}$

$n_{1} + n_{2} = 4$
$n_{2} - n_{1} = 2$
$2 n_{2} = 6$
$n_{2} = 3$
$n_{1} = 1$
$3 \to 1$
$= R Z^{2} (\frac{1}{n _{1}^{2}} - \frac{1}{n _{2}^{2}})$
$= R \times 2^{2} (\frac{1}{1 ^{2}} - \frac{1}{3 ^{2}})$
$= 4 R \times \frac{8}{9}$
$= \frac{32 R}{9}$

Q. When a transition of electron in He+takes place from n2​ to n1​ then wave number in terms of Rydberg constant R will be ( Given n1​+n2​=4,n2​−n1​=2 )

43R​

98R​

932R​

924R​