A point source of light is placed at the centre of curvature of a hemispherical surface. The source emits a power of $24\,W$ The radius of curvature of hemisphere is $10\,cm$ and the inner surface is completely reflecting. The force on the hemisphere due to the light falling on it is $..........\times 10^{-8}\,N$.

In a photo cell, the photo-electrons emission takes place

Light of intensity $10^{-5}\; W m ^{-2}$ falls on a sodium photo-cell of surface area $2 \;cm ^{2}$. Assuming that the top $5$ layers of sodium absorb the incident energy, estimate time required for photoelectric emission in the wave-picture of radiation. The work function for the metal is given to be about $ 2\; eV$. What is the implication of your answer?

The energy flux of sunlight reaching the surface of the earth is $1.388 \times 10^{3} \;W / m ^{2} .$ How many photons (nearly) per square metre are incident on the Earth per second? Assume that the photons in the sunlight have an average wavelength of $550\;nm$

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

When light of a given wavelength is incident on a metallic surface, the minimum potential needed to stop the emitted photoelectrons is $6.0 V$. This potential drops to $0.6 V$ if another source with wavelength four times that of the first one and intensity half of the first one is used. What are the wavelength of the first source and the work function of the metal, respectively? $\left[\right.$ Take $\left.=\frac{h c}{c}=1.24 \times 10^{-6} Jm ^{-1}.\right]$