Question

In a H-Atom, the \[n\] transition takes place from \[L\]to \[K\] shell. If \[R=1.08\times 1067{{m}^{-1}}\], the wavelength of light emitted is nearly:
A. \[4400{{A}^{0}}\]
B. \[1250{{A}^{0}}\]
C. \[1650{{A}^{0}}\]
D. \[1850{{A}^{0}}\]

Answer 1

Hint: To solve this question we need to first understand the Bohr’s model of hydrogen. As per Bohr's Theory, electrons of an atom rotate around the nucleus in specific orbits, or electron shells. Each orbit has its particular energy level, which is expressed as a negative value. The orbits nearer to the nucleus have lower energy levels since they interact more with the nucleus, and vice versa. Bohr named the orbits as: \[K\left( n=1 \right),L\left( n=2 \right),M\left( n=3 \right),N\left( n=4 \right),O\left( n=5 \right),....\]

Complete step by step solution: John Rydberg uses balmers work to derive an equation for all the electron transitions in the hydrogen atom.
\[R\], here is Rydberg constant
\[\lambda \], is the wavelength \[n\] is equal to the energy level( initial and final )
For \[K\] and \[L\] shell corresponding principal quantum number \[n=1\]and \[n=2\]
Transition is from \[{{n}_{2}}=2\] to \[{{n}_{1}}=1\]
\[\dfrac{1}{\lambda }=R\left[ \dfrac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}} \right]\]
Now, substituting \[{{n}_{1}}=1,{{n}_{2}}=2\] and \[R=1.09\times {{10}^{7}}{{m}^{-1}}\]
\[
  \lambda =\dfrac{4}{3R} \\
 \Rightarrow \lambda =1250{{A}^{0}} \\
\]
Hence, the correct option is B. \[1250{{A}^{0}}\].

Additional Information: The estimation of the Rydberg constant \[{{R}_{\infty }}=1.097373\times {{10}^{7}}\] for each meter. At the point when utilized in this structure in the numerical description of a series of spectral lines, the outcome is the waves of waves per unit length, or the wave numbers. Multiplication by the speed of light yields the frequencies of the spectral lines.

Note: Since the energy level of the electron of a hydrogen atom is quantized rather than consistent, the range of the lights emitted by the electron through transition is likewise quantized. All in all, the frequency can just be an interpretation of specific values since \[{{n}_{1}}\]and \[{{n}_{2}}\] are integers. Subsequently, the electron transition gives spectral lines.