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Question: In which option is the highest velocity of electrons? (A) \( H,n = 2 \) (B) \( H{e^ + },n = 6 \...

In which option is the highest velocity of electrons?
(A) H,n=2H,n = 2
(B) He+,n=6H{e^ + },n = 6
(C) Li2+,n=3L{i^{2 + }},n = 3
(D) Be3+,n=5B{e^{3 + }},n = 5

Explanation

Solution

The electron with highest kinetic energy has highest velocity, as the kinetic energy is directly proportional to the square of velocity. Kinetic energy must be in joules and can be calculated from the energy of the electron present in that shell. By substituting the atomic number and principal quantum number in the equation of En{E_n} gives the energy.

Complete answer:
Given that they are four different ions, with electrons in different shells. The value of nn represents the principal quantum number. From the ions the atomic number can be known.
The energy of electron is given by En=−13.6(Z)2n2eV{E_n} = \dfrac{{ - 13.6{{\left( Z \right)}^2}}}{{{n^2}}}eV
Substitute the atomic number in ZZ and principal quantum number in nn
The obtained energy will be in electron volts, convert this energy into joules by multiplying the value with 1.6×10−191.6 \times {10^{ - 19}} , as 1eV=1.6×10−19J1eV = 1.6 \times {10^{ - 19}}J
The energy of electron in hydrogen ion is E2(H)=−13.6(1)222=−3.403eV=5.452×10−19J{E_2}\left( H \right) = \dfrac{{ - 13.6{{\left( 1 \right)}^2}}}{{{2^2}}} = - 3.403eV = 5.452 \times {10^{ - 19}}J
The energy of electron in helium ion is E6(He+)=−13.6(2)262=−1.512eV=2.422×10−19J{E_6}\left( {H{e^ + }} \right) = \dfrac{{ - 13.6{{\left( 2 \right)}^2}}}{{{6^2}}} = - 1.512eV = 2.422 \times {10^{ - 19}}J
The energy of electron in lithium ion is E3(Li2+)=−13.6(3)232=−1.3.6eV=2.18×10−18J{E_3}\left( {L{i^{2 + }}} \right) = \dfrac{{ - 13.6{{\left( 3 \right)}^2}}}{{{3^2}}} = - 1.3.6eV = 2.18 \times {10^{ - 18}}J
The energy of electron in beryllium ion is E5(Be3+)=−13.6(4)252=−8.710eV=1.395×10−18J{E_5}\left( {B{e^{3 + }}} \right) = \dfrac{{ - 13.6{{\left( 4 \right)}^2}}}{{{5^2}}} = - 8.710eV = 1.395 \times {10^{ - 18}}J
The magnitude of energy in electrons is nothing but kinetic energy which has the below formula.
Thus, ∣En∣=K=12mv2\left| {{E_n}} \right| = K = \dfrac{1}{2}m{v^2}
Now, the velocity can be written as v=2Kmev = \sqrt {\dfrac{{2K}}{{{m_e}}}}
Substitute the obtained kinetic energies in the above formula, mass of electron as 9.1×10−31kg9.1 \times {10^{ - 31}}kg
The velocity of electron in hydrogen ion is v2(H)=2×4.52×10−199.1×10−31=1.094×106ms−1{v_2}\left( H \right) = \sqrt {\dfrac{{2 \times 4.52 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}} = 1.094 \times {10^6}m{s^{ - 1}}
The velocity of electron in helium ion is v6(He+)=2×2.422×10−199.1×10−31=7.292×105ms−1{v_6}\left( {H{e^ + }} \right) = \sqrt {\dfrac{{2 \times 2.422 \times {{10}^{ - 19}}}}{{9.1 \times {{10}^{ - 31}}}}} = 7.292 \times {10^5}m{s^{ - 1}}
The velocity of electron in lithium ion is v3(Li2+)=2×2.180×10−189.1×10−31=2.188×106ms−1{v_3}\left( {L{i^{2 + }}} \right) = \sqrt {\dfrac{{2 \times 2.180 \times {{10}^{ - 18}}}}{{9.1 \times {{10}^{ - 31}}}}} = 2.188 \times {10^6}m{s^{ - 1}}
The velocity of electron in beryllium ion is v5(Be3+)=2×1.395×10−189.1×10−31=1.750×106ms−1{v_5}\left( {B{e^{3 + }}} \right) = \sqrt {\dfrac{{2 \times 1.395 \times {{10}^{ - 18}}}}{{9.1 \times {{10}^{ - 31}}}}} = 1.750 \times {10^6}m{s^{ - 1}}
The velocity of electrons in lithium ions is more.
Electron in option C has the highest velocity.

Note:
When energy is given to an electron, it undergoes transition from one orbit to another orbit, and the energy can be expressed in different units like electron volts, or Joules. While calculating the kinetic energy, the energy of electrons must be in joules. Thus, conversion should be made from eV to joules.