$A$ sodium atom is in one of the states labeled 'Lowest excited levels'. It remains in that state for an average time of $10^{-8} \, s$,before it makes a transition back to a ground state. What is the uncertainty in energy of that excited state?

In the Davisson-Germer experiment,for which value of accelerating voltage was the intensity of the electron beam found to be maximum (in $V$)?

The wave nature of electrons suggests that they will produce diffraction effects. Davisson and Germer experimentally proved this by diffracting electrons from a crystal. The law of diffraction from a crystal is obtained from the constructive interference of electron waves reflected from the planes of atoms in the crystal. Electrons accelerated by a potential $V$ are diffracted by a crystal. If $d = 1 \ \mathring{A}$ and $i = 30^\circ$,then the value of $V$ should be ........... $V$. $(h = 6.6 \times 10^{-34} \ J \cdot s, m_e = 9.1 \times 10^{-31} \ kg, e = 1.6 \times 10^{-19} \ C)$

According to Bohr's theory,the electron in orbits has definite energy values. Then,according to the uncertainty principle,the lifetime of an excited state will be:

Statement $1$: The Davisson-Germer experiment established the wave nature of electrons. Statement $2$: If electrons possess wave nature,they can undergo interference and diffraction.

Using the Heisenberg uncertainty principle,arrange the following particles in the order of increasing lowest energy possible.
$(I)$ An electron in $H_{2}$ molecule
$(II)$ $A$ hydrogen atom in a $H_{2}$ molecule
$(III)$ $A$ proton in the carbon nucleus
$(IV)$ $A$ $H_{2}$ molecule within a nanotube