Atoms Test - 60...

Wavelength of the first line of Balmer series is $$600nm$$. The wavelength of second line of the Balmer series will be:

Ionized hydrogen atoms and $$\alpha$$-particles with same momenta enters perpendicular to a constant magnetic field, B. The ratio of their radii of their paths $$r_H : r_{\alpha}$$ will be :

The total energy of an electron revolving in the second orbit of hydrogen atom is?

The period of revolution of an electron in the ground state of hydrogen atom is T. The period of revolution of the electron in the first excited state is?

In $$H$$-atm spectrum $$V$$ is the wave number
$$V_1 = V_{min} + V_{max}$$ for Lyman series
$$V_1 = V_{min} + V_{max}$$ for Balmer series then $$V_1 : V_2$$

An excited $$He^+$$ ion emits two photons in succession, with wavelengths $$108.5 nm$$ and $$30.4 nm$$, in making a transition to ground state. The quantum number $$n$$, corresponding to its initial excited state is (for photon of wavelength $$\lambda$$, energy $$E = \dfrac{1240 eV}{\lambda (in\, nm)}$$)

The number density of molecules of a gas depends on their distance $$r$$ from the origin as, $$n(r) = n_0 e^{-\alpha r^4}$$. Then the total number of molecules is proportional to:

The ratio of mass densities of nuclei of $$^{40} Ca$$ and $$^{16} O$$ is close to :

Taking the wavelength of first Balmer line in hydrogen spectrum ($$n=3$$ to $$n=2$$) as $$660nm$$, the wavelength of the 2nd Balmer line ($$n=4$$ to $$n=2$$) will be:

In $$Li^{++}$$, electron in first Bohr orbit is excited to a level by a radiation of wavelength $$\lambda$$. when the ion gets deexcited to the ground state in all possible ways (including intermediate emissions), a total of six spectral lines are observed. What is the value of $$\lambda$$ ?
(Given : $$h = 6.63 \times 10^{34} Js$$;  $$ \, c = 3 \times 10^8 ms^{-1}$$)