Question

Figure 37.7 identified the wavelengths of four lines in the Balmer series of hydrogen.

a. Determine the Balmer formula n and m values for these wavelengths.

b. Predict the wavelength of the fifth line in the spectrum.

Short answer

(a) For $\lambda =656.6nm$: $n=3$and $m=2$

For $\lambda =486.3nm$: $n=4$and $m=2$

For $\lambda =434.2nm$: $n=5$and $m=2$

For $\lambda =410.3nm$: $n=6$and $m=2$

(b) The wavelength of the fifth line in the spectrum is$\lambda =397.14nm$.

Step-by-step solution

Part(a) Step 1: Given information

We have given that the wavelengths of the four lines in the Balmer series of hydrogen are $\lambda =656.6nm$, $\lambda =486.3nm$, $\lambda =434.2nm$and $\lambda =410.3nm$.

We need to find the Balmer formula n and m values for the given wavelengths.

Part(a) Step 2: Simplify

We know that the formula for the Balmer series is,

$\frac{1}{m^2}-\frac{1}{n^2}=\frac{91.18\left(nm\right)}{\lambda \left(nm\right)}$, where ($m=1,2,3$and role="math" localid="1649755449614" $n=m+1,m+2,...$)

On simplifying,

role="math" localid="1649757329881" $\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=\frac{91.18\left(nm\right)}{\lambda \left(nm\right)}$ ( Let this equation be ($1$) )

Now find the value of n and m which satisfy the formula of Balmer series given in equation ($1$) for each wavelength.

For $\lambda =656.6nm$:

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=\frac{91.18\left(nm\right)}{656.6\left(nm\right)}$ ( By substituting $\lambda =656.6nm$)

On simplifying,

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=0.1389$ ( Let this equation be ($2$) )

The equation ($2$) is satisfied for $n=3$and $m=2$.

For $\lambda =486.3nm$:

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=\frac{91.18\left(nm\right)}{486.3\left(nm\right)}$ ( By substituting $\lambda =486.3nm$)

On simplifying,

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=0.1875$ ( Let this equation be ($3$) )

The equation ($3$) is satisfied for $n=4$and $m=2$.

For $\lambda =434.2nm$:

role="math" localid="1649757355521" $\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=\frac{91.18\left(nm\right)}{434.2\left(nm\right)}$ ( By substituting $\lambda =434.2nm$)

On simplifying,

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=0.21$ ( Let this equation be ($4$) )

The equation ($4$) is satisfied for $n=5$and $m=2$.

For $\lambda =410.3nm$:

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=\frac{91.18\left(nm\right)}{410.3\left(nm\right)}$ ( By substituting $\lambda =410.3nm$)

On simplifying,

$\frac{n^2-m^2}{\left(m^2\right)\left(n^2\right)}=0.2222$ ( Let this equation be ($5$) )

The equation ($5$) is satisfied for $n=6$and $m=2$.

Part(b) Step 1: Given information

We have given that the wavelengths of the four lines in the Balmer series of hydrogen are

$\lambda =656.6nm$,$\lambda =486.3nm$,$\lambda =434.2nm$and $\lambda =410.3nm$.

We need to find the wavelength of the fifth line in the spectrum.

Part(b) Step 2: Simplify

The fifth line in the spectrum will correspond to $m=2$and $n=7$.

Using Balmer series formula,

$\frac{1}{m^2}-\frac{1}{n^2}=\frac{91.18\left(nm\right)}{\lambda \left(nm\right)}$

Now express the above formula in terms of $\lambda$,

$\lambda =\frac{91.18\left(nm\right)}{{\displaystyle \frac{1}{m^2}}-{\displaystyle \frac{1}{n^2}}}$

role="math" $\lambda =\frac{91.18\left(nm\right)}{{\displaystyle \frac{1}{{\left(2\right)}^2}}-{\displaystyle \frac{1}{{\left(7\right)}^2}}}$ (By substituting $m=2$and $n=7$)

$\lambda =397.14nm$