(a) Name the spectral series for a hydrogen atom which lies in the visible region. Find the ratio of the maximum to the minimum wavelengths of this series.

OR
(b) What are matter waves? A proton and an alpha particle are accelerated through the same potential difference. Find the ratio of the de Broglie wavelength associated with the proton to that with the alpha particle.

Balmer series of hydrogen region lies in visible region of electromagnetic spectrum.
When an electron in a hydrogen atom transition from a n=3 or higher orbital to a n=2 orbital, lines are released. 
According to the equation of Balmer series,
$$\begin{gathered} \frac{1}{\mathrm{\lambda }}=\mathrm{R}\left(\frac{1}{{{\mathrm{n}}^2}_1}-\frac{1}{{{\mathrm{n}}^2}_2}\right) \\ ={\mathrm{\lambda }}_{\min }/{\mathrm{\lambda }}_{\max } \\ =(1/2^2-1/3^2)/(1/2^2-1/\infty ) \\ =5/9 \end{gathered}$$
Hence, the ratio of minimum to maximum wavelength in the Balmer series is 5: 9.

OR 
The wave associated with each moving particle is known as a matter wave.
The matter-wave has a wavelength of λ = hp,
where h is the Planck’s constant and p is the moment of the moving particle.
Using formula $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$, 
where λ is de Broglie wavelength, h is Plank's constant, m is the mass of the particle, and E is the
energy of the particle.
m_p and q_p are the mass and charge of the proton respectively.
m$\mathrm{\alpha }$ and q$\mathrm{\alpha }$ is the mass and charge of the alpha particle respectively.
Energy (E) is the same of both:
$4{\mathrm{m}}_{\mathrm{p}}={\mathrm{m}}_{\mathrm{\alpha }} \mathrm{and} 2{\mathrm{q}}_{\mathrm{p}}={\mathrm{q}}_{\mathrm{\alpha }}$
Since $\mathrm{\lambda }=\frac{\mathrm{h}}{\sqrt{2\mathrm{mE}}}$
$\frac{{\mathrm{\lambda }}_{\mathrm{p}}}{{\mathrm{\lambda }}_{\mathrm{\alpha }}}=\sqrt{\frac{{\mathrm{m}}_{\mathrm{\alpha }}}{{\mathrm{m}}_{\mathrm{p}}}}=\sqrt{(}4{\mathrm{m}}_{\mathrm{p}}/{\mathrm{m}}_{\mathrm{p}})=2$
Ratio- 2:1