Question

calculate the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom?

Answer 1

`\Delta E=E_(final)-E_(initial)`

`\Delta E=-1312[(1)/((n_f^2))-\frac {1}{(n_i^2 )}]KJ mol^(-1)`

`\Delta E=-1312[(1)/(3^2))-\frac {1}{(1^2 )}]KJ mol^(-1)`

`\Delta E=-1312[(1)/((9))-\frac {1}{(1 )}]KJ mol^(-1)`

`\Delta E=-1312[0.111-1]KJ mol^(-1)`

`\Delta E=1166 KJ mol^(-1)`

`\frac{=1166,000 \mathrm{J}}{6.022 * 10^(23) \text { photons }}`

`=193623 * 10^(-23)  \frac {J}{photon}`

`\Delta E=1.93623 * 10^(-18)  \frac {J}{photon}`

`\Delta E=\frac {h* c}{\lambda} \\\\=\frac {(6.626* 10^(-34) J s * 3 * 10^8 ms^(-1))}{\lambda}`

h is planck's constant

c is the speed of light

λ is the wavelength of light

`\lambda =\frac {h* c}{\Delta E}\\\\=\frac {(6.626*10^(-34) J s*3 * 10^8 ms^(-1))}{(1.93623*10^(-18)  J/photon)}`

Wavelength

`\lambda =10.3 * 10^(-8) m * \frac {(10^9 nm)}{1m}  =103 nm (Answer)`

Thus, the wavelength of light associated with the transition from n=1 to n=3 in the hydrogen atom is 103 nm.