According to de-Broglie, the de-Broglie wavelength for an electron in an orbit of a hydrogen atom is $10^{-9} \, m$. The principal quantum number for this electron is:

The ratio of the total energy of the $2^{\text{nd}}$ orbit electron for the hydrogen atom $(_1H^1)$ to that of the helium ion $(He^+)$ $(_2^4He)$ is:

The frequency of light emitted,when the electron makes a transition from the level of principal quantum number $n=2$ to the level with $n=1$ is (Take,the ionization energy of hydrogen to be $13.6 \ eV$ and $h \simeq 4 \times 10^{-15} \ eV \cdot s$)

If $\lambda_1$ and $\lambda_2$ denote the wavelengths of de Broglie waves for electrons in the first and second Bohr orbits in a hydrogen atom,then $\lambda_1/\lambda_2$ is equal to

If an electron jumps from the $1^{st}$ orbital to the $3^{rd}$ orbital,then it will:

Consider an electron in the $n=3$ orbit of a hydrogen-like atom with atomic number $Z$. At absolute temperature $T$,a neutron having thermal energy $k_B T$ has the same de Broglie wavelength as that of this electron. If this temperature is given by $T = \frac{Z^2 h^2}{\alpha \pi^2 a_0^2 m_N k_B}$,(where $h$ is the Planck's constant,$k_B$ is the Boltzmann constant,$m_N$ is the mass of the neutron and $a_0$ is the first Bohr radius of hydrogen atom),then the value of $\alpha$ is $....$