An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ, energy E = (1240 eV/λ (in nm)) ;

Question

An excited He+ ion emits two photons in succession, with wavelengths 108.5 nm and 30.4 nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ, energy E = (1240 eV/λ (in nm)) ; (A) n = 5 (B) n = 4 (C) n = 6 (D) n = 7

Tardigrade Admin · Accepted Answer

(1/λ) = R ((1/m2)- (1/n 2))z2 (1/1085) =R ((1/m2) - (1/n2))22 (1/304) =R ((1/12) - (1/m2))22 ∴ m = 2 ∴ n = 5

Q. An excited $H e^{+}$ ion emits two photons in succession, with wavelengths $108.5 nm$ and $30.4 nm$ , in making a transition to ground state. The quantum number $n$ , corresponding to its initial excited state is (for photon of wavelength $λ$ , energy $E = \frac{1240 e V}{λ ( in nm )}$ ;

$n = 5$

$n = 4$

$n = 6$

$n = 7$

$\frac{1}{λ} = R (\frac{1}{m ^{2}} - \frac{1}{n ^{2}}) z^{2}$
$\frac{1}{1085} = R (\frac{1}{m ^{2}} - \frac{1}{n ^{2}}) 2^{2}$
$\frac{1}{304} = R (\frac{1}{1 ^{2}} - \frac{1}{m ^{2}}) 2^{2}$
$∴ m = 2$
$∴ n = 5$

Q. An excited He+ ion emits two photons in succession, with wavelengths 108.5nm and 30.4nm, in making a transition to ground state. The quantum number n, corresponding to its initial excited state is (for photon of wavelength λ, energy E=λ(innm)1240eV​ ;

n=5

n=4

n=6

n=7