Question #103273
What is the value of ni for an electron that emits a photon of wavelength 93.85 nm when it returns to the ground state in the H atom?
1
Expert's answer
2020-02-18T06:38:56-0500

It is given that electrons emits a photon of wavelength 93.85 nm and then returns to ground state from ni the energy state, λ=93.85×107cm\lambda = 93.85 \times 10^-7 cm

now using Rydberg formula for wavelength of photon ejected

we know that

1λ\frac{1}{\lambda} =Rh(1nf21ni2)R_h(\frac{1}{n_f^2} -\frac{1}{n_i^2})

here, λ\lambda = wavelength of photon ejected, ni=n_i = energy level from which electron coming and nfn_f = energy level at which it is going=1

RhR_h = Rydberg constant = 109737.32 cm-1

now ,

1λ=109737.32(1121ni2)\frac{1}{\lambda} = 109737.32 ( \frac{1}{1^2} - \frac{1}{n_i^2} )


193.85×107=109737.32(11ni2)\frac{1}{93.85 \times10^-7} = 109737.32 (1- \frac{1}{n_i^2})


.010655 ×107=109737.32(11ni2)\times 10^7 =109737.32 (1-\frac{1}{n_i^2})


1.06×105109737.32=(11ni2)\frac{1.06 \times10^5}{109737.32 } =(1- \frac{1}{n_i^2})


.97098= 1 1ni2-\frac{1}{n_i^2}

1ni2=1.97098\frac{1}{n_i^2} = 1-.97098


ni2=1.029n_i^2 = \frac{1}{.029}


ni2=34.46n_i^2= 34.46


nin_i =5.87

so here we take minimum energy level as nin_i =6


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