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This is one of our many quizzes on Quantum mechanics. Click on this to access more such competitive quizzes on Modern Physics and Semiconductors.

Quantum mechanics Quiz 2

physics

 Two electrons are revolving around a nucleus of a hydrogen atom. If one of them is revolving in the third orbit and the other in the fourth orbit, determine the minimum possible distance in Angstrom between the two electrons if the atomic number of the atoms is $Z$.

The two electrons must be on the same side with respect to the nucleus. The minimum distance can be calculated by calculating the radii of the two concerned orbits. 

  $\text{The minimum distance between two electrons is only possible when} \\ \text{both the electrons are on the same side of the nucleus}\text{.} \\ \text{In other words the nucleus is not in between the two electrons}\text{.} \\ \text{The radius of the }n\text{-th orbit is given by;} \\ {{r}_{n}}=\dfrac{{{r}_{0}}{{n}^{2}}}{Z} \\ \Rightarrow {{r}_{n}}=\dfrac{0.529{{n}^{2}}}{Z}\text{ }\overset{\circ }{\mathop{\text{A}}}\, \\ \text{The minimum possible distance between two electrons is the difference in the} \\ \text{radii of the orbits they revolve in}\text{.} \\ \text{The minimum possible distance is given by;} \\ \Rightarrow \Delta ={{r}_{4}}-{{r}_{3}} \\ \Rightarrow \Delta =\dfrac{0.529\times {{4}^{2}}}{Z}-\dfrac{0.529\times {{3}^{2}}}{Z} \\ \Rightarrow \Delta =\dfrac{0.529}{Z}\left( 16-9 \right)=\dfrac{0.529\times 7}{Z} \\ \therefore \Delta =\dfrac{3.703}{Z}\text{ }\overset{\circ }{\mathop{\text{A}}}$

 Refer to the following energy diagram for a hypothetical hydrogen atom and find the value of $Z$ and the potential energy of the electron in the first orbit?<br><img  src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/aaa02be9-37d6-44df-8e42-fcc3c1deab9d8596586127253405635.png"/> 

 The energy of an electron in a particular orbital of a certain atom is directly proportional to the ratio of the squares of the atomic number and the orbital number. 

  $\text{Energy of electron in any orbit is given as;} \\ {{E}_{n}}=-13.6\dfrac{{{Z}^{2}}}{{{n}^{2}}} \\ \text{For }n=1\text{ ;} \\ \Rightarrow {{E}_{1}}=-13.6{{Z}^{2}}=-54.5\text{ eV} \\ \Rightarrow {{Z}^{2}}=\dfrac{54.5}{13.6} \\ \therefore Z=2 \\ \text{The potential energy of an electron is given by;} \\ U=2E \\ \Rightarrow U=2\times \left( -54.5 \right) \\ \Rightarrow U=-108.8\text{ eV} \\ \Rightarrow U=108.8\times 1.6\times {{10}^{-19}}\text{ J} \\ \therefore U=1.74\times {{10}^{-17}}\text{ J}$

 For a certain photocell if a monochromatic incident light of wavelength $\lambda $ results in the maximum velocity of the photoelectron to be $v$ , then what is the minimum potential difference required for the emission of photoelectrons? 

 The minimum potential difference required for the photoelectric effect to take place is the voltage corresponding to the threshold wavelength. 

 &nbsp;$\text{The Einstein }\!\!'\!\!\text{ s equation is given as;} \\ {{K}_{\max }}=\dfrac{hc}{\lambda }-\dfrac{hc}{{{\lambda }_{0}}} \\ \Rightarrow \dfrac{m{{v}^{2}}}{2}=hc\left( \dfrac{1}{\lambda }-\dfrac{1}{{{\lambda }_{0}}} \right) \\ \Rightarrow \dfrac{1}{{{\lambda }_{0}}}=\dfrac{1}{\lambda }-\dfrac{m{{v}^{2}}}{2hc} \\ \Rightarrow \dfrac{1}{{{\lambda }_{0}}}=\dfrac{2hc-\lambda m{{v}^{2}}}{2hc\lambda } \\ \Rightarrow {{\lambda }_{0}}=\dfrac{2hc\lambda }{2hc-\lambda m{{v}^{2}}} \\ \text{For electron, wavelength is given by;} \\ \lambda =\dfrac{12.27}{\sqrt{V}}\text{ }\overset{\circ }{\mathop{\text{A}}}\, \\ \Rightarrow {{\lambda }_{0}}=\dfrac{12.27}{\sqrt{{{V}_{0}}}} \\ \Rightarrow \sqrt{{{V}_{0}}}=\dfrac{12.27}{{{\lambda }_{0}}}=\dfrac{12.27\left( 2hc-\lambda m{{v}^{2}} \right)}{2hc\lambda } \\ \therefore {{V}_{0}}=\dfrac{150.55{{\left( 2hc-\lambda m{{v}^{2}} \right)}^{2}}}{4{{h}^{2}}{{c}^{2}}{{\lambda }^{2}}}$

 For an atom of molybdenum $Z=42$ , an electron jumps from $n=1$ to $n=2$ , the X-Ray radiation due to the jump has wavelength $\lambda =\dfrac{4}{5043Rc}$ , determine which lines does this X-ray belong. 

 X-Ray is a highly energetic electromagnetic radiation which obeys the Mosley’s law. 

 $\text{Given that;} \\ \lambda =\dfrac{4}{5043Rc} \\ \text{Wavenumber is given by;} \\ \upsilon =\dfrac{1}{\lambda }=\dfrac{5043Rc}{4} \\ \text{According to Mosley }\!\!'\!\!\text{ s Law;} \\ \sqrt{\upsilon }=a\left( Z-b \right) \\ \text{The constant }a\text{ is given as;} \\ a=\sqrt{Rc\left( \dfrac{1}{n_{l}^{2}}-\dfrac{1}{n_{2}^{2}} \right)}=\sqrt{Rc\left( \dfrac{1}{1}-\dfrac{1}{4} \right)}=\sqrt{\dfrac{3Rc}{4}} \\ \text{Substituting the values;} \\ \Rightarrow \sqrt{\dfrac{5043Rc}{4}}=\left( \sqrt{\dfrac{3Rc}{4}} \right)\left( 42-b \right) \\ \Rightarrow \sqrt{1681}=42-b \\ \Rightarrow 41=42-b \\ \therefore b=42-41=1 \\ \text{Therefore, the radiation is a }{{K}_{\alpha }}\text{ X-ray radiation}\text{.}$

 Two electrons $A$ and $B$ are revolving in two different orbits such that the maximum possible distance between the electrons is $10R$ and the minimum possible distance is $2R$ , if the atomic number of the atom is $4$ , determine which orbits do these electrons belong if $A$ is closer to nucleus than $B$. 

 The longest possible distance between two electrons revolving in different orbits of the same atom is the sum of the radii of the two orbits whereas the smallest possible distance is the difference between the radii of the orbits. 

 $\text{The minimum distance between two electrons is only possible if they are on the same} \\ \text{side of the nucleus}\text{.} \\ \text{In that case the minimum possible distance is the difference between the radii of the} \\ \text{orbits they are revolving in}\text{.} \\ \text{Therefore, according to the given condition;} \\ {{R}_{B}}-{{R}_{A}}=2R \\ \Rightarrow {{R}_{B}}=2R+{{R}_{A}}\text{                                               }\to \left( 1 \right) \\ \text{The maximum distance between two electrons is only possible if the electrons are on} \\ \text{opposite sides of the nucleus, i}\text{.e}\text{. the nucleus is between the two electrons}\text{.} \\ \text{In that case the maximum possible distance is the sum of the radii of the orbits the} \\ \text{electrons are revolving in}\text{.} \\ \text{Therefore, according to the question;} \\ {{R}_{A}}+{{R}_{B}}=10R \\ \text{From equation 1;} \\ \Rightarrow {{R}_{A}}+2R+{{R}_{A}}=10R \\ \Rightarrow 2{{R}_{A}}=8R \\ \Rightarrow {{R}_{A}}=4R \\ \Rightarrow \dfrac{0.529n_{A}^{2}}{4}=4\left( \dfrac{0.529}{4} \right)\text{                                     }\left( \because {{R}_{1}}=R=\dfrac{0.529}{4} \right) \\ \Rightarrow n_{A}^{2}=4 \\ \therefore {{n}_{A}}=2 \\ \text{Also,} \\ {{R}_{B}}=2R+{{R}_{A}}=2\left( \dfrac{0.529}{4} \right)+\left( \dfrac{0.529n_{A}^{2}}{4} \right) \\ \Rightarrow \dfrac{0.529n_{B}^{2}}{4}=2\left( \dfrac{0.529}{4} \right)+\left( \dfrac{0.529\times 4}{4} \right) \\ \Rightarrow n_{B}^{2}=2+4=6 \\ \therefore {{n}_{B}}=\sqrt{6}=2.449\approx 3$