
Estimate the difference in energy between the first and second Bohr orbit for the hydrogen atom. At what minimum atomic number would a transition from n = 2 to n = 1 energy level result in the emission of X-rays with $\lambda$ = 3.0 $\times$ 10$^{-8}$ m? Which hydrogen-like species does this atomic number correspond to?
Answer
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Hint: Use the formula of difference in energy, and wavelength can be substituted in the Bohr orbit equations, then we can determine the atomic number and the corresponding species.
Complete answer:
First, we are given with the transition of levels, i.e. n = 2 to n = 1. So, let us say that n$_1$ =1, and n$_2$ = 2.
As we know, $\Delta$E = h$\nu$ = $\dfrac{hc}{\lambda}$, here h is Planck’s constant (6.626 $\times$ 10$^{-34}$), c is the speed of light, and $\lambda$ is wavelength.
Now, wave number in the terms of wavelength, i.e.$\dfrac{1}{\lambda}$ = R[$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$], R is Rydberg’ s constant, and the value of R is 109677 cm$^{-1}$
Thus, substitute the value of $\lambda$ in the $\Delta$E = h$\nu$ = $\dfrac{hc}{\lambda}$, we get $\Delta$E = R.h.c [$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$] ,
By solving, $\Delta$E =1.635 $\times$ 10$^{-18}$ J, as $\lambda$ = 3.0 $\times$ 10$^{-8}$ m
For hydrogen like species $\Delta$E = Z$^{2}$ R.h.c [$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$]
$\dfrac{1}{\lambda}$ = Z$^{2}$R[$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$], so substitute all the values, and Z represents the atomic number.
$\dfrac{1}{3.0\times10^{-8}}$= Z$^{2}$ $\times$ 1.09678 $\times$ 10$^{7}$ [$\dfrac{1}{ 1^{2}}$ - $\dfrac{1}{2^{2}}$]
By solving we get Z$^{2}$ = 4,
Therefore, Z = 2.
So, we can say that the minimum atomic number is 2 at which the transition n = 2 to n = 1 would take place.
Atomic number 2 corresponds to the He$^{+}$, the hydrogen-like species; as the Bohr’s theory is applicable for mono-electronic species.
Note: Don’t get confused between the terms like wavenumber, and wavelength. The wavenumber shows its dependence on the wavelength. For the hydrogen atom Z = 1, but for the hydrogen-like ion, or species Z > 1. It is only applicable for hydrogen, or hydrogen-like species because it considers the interactions between one electron and the nucleus.
Complete answer:
First, we are given with the transition of levels, i.e. n = 2 to n = 1. So, let us say that n$_1$ =1, and n$_2$ = 2.
As we know, $\Delta$E = h$\nu$ = $\dfrac{hc}{\lambda}$, here h is Planck’s constant (6.626 $\times$ 10$^{-34}$), c is the speed of light, and $\lambda$ is wavelength.
Now, wave number in the terms of wavelength, i.e.$\dfrac{1}{\lambda}$ = R[$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$], R is Rydberg’ s constant, and the value of R is 109677 cm$^{-1}$
Thus, substitute the value of $\lambda$ in the $\Delta$E = h$\nu$ = $\dfrac{hc}{\lambda}$, we get $\Delta$E = R.h.c [$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$] ,
By solving, $\Delta$E =1.635 $\times$ 10$^{-18}$ J, as $\lambda$ = 3.0 $\times$ 10$^{-8}$ m
For hydrogen like species $\Delta$E = Z$^{2}$ R.h.c [$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$]
$\dfrac{1}{\lambda}$ = Z$^{2}$R[$\dfrac{1}{ n_1^{2}}$ - $\dfrac{1}{ n_2^{2}}$], so substitute all the values, and Z represents the atomic number.
$\dfrac{1}{3.0\times10^{-8}}$= Z$^{2}$ $\times$ 1.09678 $\times$ 10$^{7}$ [$\dfrac{1}{ 1^{2}}$ - $\dfrac{1}{2^{2}}$]
By solving we get Z$^{2}$ = 4,
Therefore, Z = 2.
So, we can say that the minimum atomic number is 2 at which the transition n = 2 to n = 1 would take place.
Atomic number 2 corresponds to the He$^{+}$, the hydrogen-like species; as the Bohr’s theory is applicable for mono-electronic species.
Note: Don’t get confused between the terms like wavenumber, and wavelength. The wavenumber shows its dependence on the wavelength. For the hydrogen atom Z = 1, but for the hydrogen-like ion, or species Z > 1. It is only applicable for hydrogen, or hydrogen-like species because it considers the interactions between one electron and the nucleus.
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