The prominent yellow line in the spectrum of sodium vapour lamp has a wave length of $590 \mathrm{nm}$. What minimum accelerating potential is needed to excite this line in an electron tube containing sodium vapour? $\left(\mathrm{h}=6.63\times {10}^{-34} \mathrm{J}·\mathrm{s}, \mathrm{c}=3\times {10}^8 \mathrm{m} {\mathrm{s}}^{-1}\right)$

Round off your answer to the nearest integer.

If the energy difference between the ground state of an atom and its excited state is $4.4\times {10}^{-19} \mathrm{J}$, The wave length of the photon (in meters) required to produce this transition is $\frac{\mathrm{x}}{2}\times {10}^{-7}\mathrm{m}$. The value of $\mathrm{x}$ is

Round off your answer up to the nearest integer.

Electromagnetic radiation of wavelength $242 \mathrm{nm}$ is just sufficient to ionise a sodium atom. Calculate the ionization energy of sodium atom $\left(\mathrm{in} \mathrm{kJ}/\mathrm{mole}\right)$. $\mathrm{c}=3\times {10}^8 {\mathrm{ms}}^{-1}$, $\mathrm{h}=6.626\times {10}^{-34} \mathrm{Js}.$

Report your answer to the nearest integer.

Calculate the wavelength in nanometer associated with a proton moving at $1.0\times {10}^3\mathrm{meter} {\mathrm{second}}^{-1}. \left(\text{Mass of proton }=1.67\times {10}^{-27}\mathrm{kilogram}\mathrm{and}\mathrm{h}=6.63\times {10}^{-34}\mathrm{Joule} \mathrm{second}\right)$

In an atom, an electron is moving with a speed of $600 \mathrm{m}/\mathrm{s}$ with an accuracy of $0.005\%$. Certainty with which the position of the electron can be located is,$\left(\mathrm{h}=6.6\times {10}^{-34} \mathrm{kg} {\mathrm{m}}^2{\mathrm{s}}^{-1}\mathrm{,} {\mathrm{m}}_{\mathrm{e}}=9.1\times {10}^{-31} \mathrm{kg}\right)$