Find the radius of the n = 4 Bohr orbit of a doubly ionized lithium atom (Li2+,

Question

Find the radius of the n = 4 Bohr orbit of a doubly ionized lithium atom (Li2+, Z = 3). Is the energy required to raise an electron from the n = 4 state to the n = 6 state in Li2+ greater than, less than, or equal to the energy required to raise an electron in hydrogen from the state to the n = 6 state? |Delta Eion| |Delta E hI|" greater force on Li2+ electron due to larger number of protons |Delta Eionn| = |Delta E h| " same n interval in two hydrogen-like atoms Verify your answer to part by calculating the relevant energies. Delta ELi2+ = j Delta EH = J

Explanation / Answer

a)

Here , for Li2+ ion

atomic number , Z = 3

as the bohr radius is given as

Rn = n^2 * 0.0529 nm/Z

for n = 4

Rn = 4^2 * 0.0529 *10^-9/3

Rn = 2.821 *10^-10 m

the radius of orbit is 2.821 *10^-10 m

b)

the energy needed to raise electron from n1 to n2 is given as

E = 13.6 * Z^2 * (1/n1^2 - 1/n2^2)

as Z is greater for Li2+ than H

the correct option is 4

delta Eion > delta EH --- greater force on Li2+ due to large number of protons

c)

for Li2+

delta Eion = 13.6 * Z^2 * (1/n1^2 - 1/n2^2)

delta Eion = 13.6 * 3^2 * (1/4^2 - 1/6^2) eV

delta Eion = 4.25 eV = 4.25 * 1.602 *10^-19 J

delta Eion = 6.81 *10^-19 J

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for Hydrogen

delta EH = 13.6 * (1/n1^2 - 1/n2^2)

delta EH = 13.6 * (1/4^2 - 1/6^2) eV

delta EH = 0.47 eV = 0.47 * 1.602 *10^-19 J

delta EH = 7.56 *10^-20 J

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