Question

Select the correct statements for shifting all the electrons from first Bohr orbit to the sixth Bohr orbit in $1$ mole of hydrogen:

• A. Energy required in the above process is $21.12\times {10}^{-19}{\text{J atom}}^{-1}$
• B. The distance that an electron have to cover would be $18.5\stackrel{o}{A}$
• C. Energy required in the above process per mole is $6.48\times {10}^4\text{cal}$
• D. Energy required in the above process per mole is $12.67\times {10}^5{\text{J mol}}^{-1}$

Answer 1

Answer: (A), (B), (D)

Energy required in the above process per mole is $12.67\times {10}^5{\text{J mol}}^{-1}$

The energy required to shift electrons from first Bohr orbit to sixth Bohr orbit in hydrogen atom $Z=1$ can be calculated by using the relation,
$\Delta E=13.6(\frac{1}{n_1^2}-\frac{1}{n_2^2})$
Substituting the values, we get
$\Delta E=13.6(\frac{1}{1^2}-\frac{1}{6^2})\text{eV}$
$\Delta E=13.6\times \frac{35}{36}\text{eV}$
$=13.2\times 1.6\times {10}^{-19}{\text{J atom}}^{-1}=21.12\times {10}^{-19}{\text{J atom}}^{-1}$
$\therefore \Delta E/mol=21.12\times {10}^{-19}\times 6\times {10}^{23}=12.67\times {10}^5{\text{J mol}}^{-1}$

The distance that each electron have to cover in this transition would be $=\Delta r=r_6-r_1$
where,
$r_6=\text{Radius of 6th orbit}$
$r_1=\text{Radius of 1st orbit}$
$\therefore \Delta r=0.529(6^2-1^2)\stackrel{o}{A}=(0.529\times 35)\stackrel{o}{A}=18.515\stackrel{o}{A}$