Question

Question 16 5 pts Use the Bohr model to find the second longest wavelength of light in the Balmer series for a triply-ionized Be atom (Z = 4). Recall that the Balmer series corresponds to transitions to the n = 2 level. 30.4 nm 117 nm 73.0 nm 41.1 nm 209 nm

Answer 1

The formula for the wavelength $\lambda$ of the emitted radiation from an atom is given by,

$\frac{1}{\lambda}=RZ^{2}(\frac{1}{n{_{1}}^{2}}-\frac{1}{n{_{2}}^{2}})$

where R is the Rydberg's constant (R = 1.097 x 10^7m^-1) and Z is the atomic number.

For balmer series n1 = 2 and for the secons longest wavelength n2 = 4. Also for Be atom Z = 4. Therefore,

$\frac{1}{\lambda}=R\times 4^{2}(\frac{1}{2^{2}}-\frac{1}{4^{2}})$

$\frac{1}{\lambda}=R\times 16(\frac{1}{4}-\frac{1}{16})$

$\frac{1}{\lambda}=R\times 16\times \frac{12}{4\times 16}$

$\frac{1}{\lambda}=3R$

$\lambda=\frac{1}{3R}=\frac{1}{3\times 1.097\times 10^{7}}$

$\lambda=30.4\times 10^{-9}m=30.4nm$

So the second largest wavelength of the light in the Balmer series for a triply-ionised Be atom is 30.4nm.

First option is the correct answer.