The shortest wavelength of a photon that can be emitted by a hydrogen atom,for which the initial state is $n = 12 ,$ is closest to which one of the following values? $\left( c = 3.0 \times 10 ^ { 8 } \mathrm {~m} / \mathrm { s } , \right.$ $h = 6.626 \times 10 ^{- 34} \mathrm {~J} \cdot \mathrm { s }$ $1 \mathrm { eV } = 1.60 \times 10^{ - 19} \mathrm {~J} )$
A)92 nm
B)82 nm
C)72 nm
D)62 nm
E)52 nm

Question

Quizplus · Accepted Answer

The shortest wavelength corresponds to the largest energy difference between the levels, which is for $n_f = 1$. The energy difference is:$$\Delta E = E_f - E_i = -13.6 \left(\frac{1}{1^2} - \frac{1}{12^2}\right) \mathrm{eV} = 10.2 \mathrm{eV}$$The wavelength is then:$$\lambda = \frac{hc}{\Delta E} = \frac{(6.626 \times 10^{-34} \mathrm{J} \cdot \mathrm{s})(3.0 \times 10^8 \mathrm{m/s})}{10.2 \mathrm{eV}(1.60 \times 10^{-19} \mathrm{J/eV})} = 1.21 \times 10^{-7} \mathrm{m} = 121 \mathrm{nm}$$The closest choice is 92 nm.

The Shortest Wavelength of a Photon That Can Be Emitted n=12,n = 12 ,n=12,

The Shortest Wavelength of a Photon That Can Be Emitted $n = 12 ,$