Physics Test - ...

If ${\lambda }_0$ is the de-Broglie wavelength for a proton accelerated through a potential difference of $100V$, the de-Broglie wavelength for $\alpha$-particle accelerated through the same potential difference is:

A muonic hydrogen atom is a bound state of a negatively charged muon (denoted by ${\mu }^{-}$) of mass \(207 m_{e}\) and a proton, and the moon orbits around the proton. What is the radius of its first Bohr orbit and its ground state energy, respectively?

The first member of the Balmer series of hydrogen spectrum has a wavelength $6563 \AA$ Compute the wavelength of the second member.

Which energy state of triply ionized beryllium $\left(B{e}^{+++}\right)$ has the same electron orbital radius as that of the ground state of hydrogen? Given $Z$ for beryllium $= 4$.

The wavelength of the first member of the Balmer series in the hydrogen spectrum is $6563 \AA$. Calculate the wavelength of the first member of the Lyman series in the same spectrum.

The specific charge of a proton is $9.6 \times {10}^7 C/kg$. The specific charge of an $\alpha$ particle will be:

A transformer:

A proton and an \(α\)-particle are accelerated through the same potential difference. The ratio of the de-Broglie wavelength of the proton to the de-Broglie wavelength of the alpha particle will be:

An electron is accelerated from rest to potential \(V\). The final velocity of the electron is:

If \(\lambda_{1}\) and \(\lambda_{2}\) are the wavelengths of the first members of Lyman and Paschen series, respectively, then \(\lambda_{1}: \lambda_{2}\) is: