If on transition to the ground state an He+ ion emits two photons in succession, having wavelengths 1026.7 overset° A and 304 overset° A , then the quantum number n corresponding to the exciting state of He+ ion is [ R=1.096× 107m- 1 ]

Question

If on transition to the ground state an He+ ion emits two photons in succession, having wavelengths 1026.7 overset° A and 304 overset° A , then the quantum number n corresponding to the exciting state of He+ ion is [ R=1.096× 107m- 1 ] (A) 4 (B) 6 (C) 2 (D) 1

Tardigrade Admin · Accepted Answer

(h c/λ1)+(h c/λ2)= operatornameRch Z2((1/12)-(1/n2)) Put λ 1=1026.7â« and λ 2=304 â« Z=2 for He+ ion On solving for n n=6

Q. If on transition to the ground state an $H e^{+}$ ion emits two photons in succession, having wavelengths $1026.7 A^{\circ}$ and $304 A^{\circ}$ , then the quantum number $n$ corresponding to the exciting state of $H e^{+}$ ion is [ $R = 1.096 \times 1 0^{7} m^{- 1}$ ]

4

6

2

1

$\frac{h c}{λ _{1}} + \frac{h c}{λ _{2}} = Rch Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{n ^{2}})$
Put $λ_{1} = 1026.7Å$ and $λ_{2} = 304 Å$
$Z = 2$ for $H e^{+}$ ion
On solving for $n$
$n = 6$

Q. If on transition to the ground state an He+ ion emits two photons in succession, having wavelengths 1026.7A∘ and 304A∘ , then the quantum number n corresponding to the exciting state of He+ ion is [ R=1.096×107m−1 ]

4

6

2

1

λ1​hc​+λ2​hc​=RchZ2(121​−n21​) Put λ1​=1026.7Å and λ2​=304Å Z=2 for He+ ion On solving for n n=6

Q. If on transition to the ground state an $H e^{+}$ ion emits two photons in succession, having wavelengths $1026.7 A^{\circ}$ and $304 A^{\circ}$ , then the quantum number $n$ corresponding to the exciting state of $H e^{+}$ ion is [ $R = 1.096 \times 1 0^{7} m^{- 1}$ ]

$\frac{h c}{λ _{1}} + \frac{h c}{λ _{2}} = Rch Z^{2} (\frac{1}{1 ^{2}} - \frac{1}{n ^{2}})$
Put $λ_{1} = 1026.7Å$ and $λ_{2} = 304 Å$
$Z = 2$ for $H e^{+}$ ion
On solving for $n$
$n = 6$