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QUESTION

# An excited hydrogen atom emits light with a wavelength of 397.2nm to reach the energy level for which n=2. In which principle quantum level did the electron did begin?

DeltaE=-R_H*(Z)^2[1/n_f^2-1/n_i^2]

In the hydrogen atom, the energy of the electron in a given energy level is given by : E_n= -R_H*(Z/n)^2

E_f= -R_H*(Z/n_f)^2

E_i= -R_H*(Z/n_i)^2

DeltaE=E_f-E_i

DeltaE=[-R_H*(Z/n_f)^2]-[ -R_H*(Z/n_i)^2]

take  -R_H*(Z)^2 as a common factor,

DeltaE=-R_H*(Z)^2[1/n_f^2-1/n_i^2]

The energy of the photon emitted is given by:

DeltaE=-hc/lambda

please note that a negative sign must be introduced to the energy expression since energy is released.

combining the two equations gives:

-hc/lambda=-R_H*(Z)^2[1/n_f^2-1/n_i^2]  (equation 1)

h" is Planck's constant" = 6.626*10^-34 J.s  R_H" is Rydberg constant" = 2.178*10^-18 J  Z" is the atomic number of the hydrogen atom" = 1 n" is principle quantum number"

n_i " is the initial quantum state of the electron." n_f =2

plugging the numbers in (equation 1)

-(6.626*10^-34 J.s*2.998*10^8 m/s) /(397.2*10^-9 m)=-2.178*10^-18 J*(1)^2[1/2^2-1/n_i^2]

-(6.626*10^-34 cancel(J).cancel(s)*2.998*10^8cancel(m)/cancel(s)) /(397.2*10^-9 cancel(m))=-2.178*10^-18 cancel(J)*(1)^2[1/2^2-1/n_i^2]

solve for n_i,

n_i= 7