Question

According to the Bohr model of the atom, which electron transition will emit the lowest energy photon?

• A. $n=2$ to $n=5$
• B. $n=3$ to $n=2$
• C. $n=3$ to $n=4$
• D. $n=6$ to $n=2$

Answer 1

Answer: (B)

$n=3$ to $n=2$

According to Bohr's model, electronic transition for 1-electron systems from state $n_1$ to $n_2$ is given by:

$\Delta E=-R{Z}^2(\frac{1}{n_2^2}-\frac{1}{n_1^2})$

For emission of photon $n_2>n_1$ i.e. for $3\to 2$

and $6\to 2$ and the other transition i.e. $2\to 5$

and $3\to 4$ will occur by absorption of a photon.

For transition of $3\to 2$:

$\Delta E=-R{Z}^2(1/2^2-1/3^2)$

$\Delta E=-R{Z}^2\times 0.14$

For transition of $6\to 2$:

$\Delta E=-R{Z}^2(1/2^2-1/6^2)$

$\Delta E=-R{Z}^2\times (-0.22)$

For transition of $2\to 5$:

$\Delta E=-R{Z}^2(1/2^2-1/5^2)$

$\Delta E=-R{Z}^2\times 0.21$

For transition of $3\to 4$:

$\Delta E=-R{Z}^2(1/3^2-1/4^2)$

$\Delta E=-R{Z}^2\times 0.048$

It is minimum for $3\to 2$ transition

Hence, option $B$ is correct