Match the following

	COLUMN-I		COLUMN-II
I)	de Broglie wavelength of electron moving with velocity half of speed of light is equal to that of photon. Then ratio $\frac{K{E}_e}{K{E}_{photon}}$ is	P)	$>1$
II)	Particle of mass M at rest decays into two particles of masses $m_1$  and  $m_2$ with non-zero velocities.  The ratio of de Broglie wavelength  $\frac{{\lambda }_1}{{\lambda }_2}$ is	Q)	$\frac{7}{8}$
III)	Proton and electron accelerated by same potential. If ${\lambda }_e$  and  ${\lambda }_p$ denote the de Broglie wavelengths of electron and proton then $\frac{{\lambda }_e}{{\lambda }_p}$  is	R)	1
IV)	Ultraviolet light of wavelengths  ${\lambda }_1$  and  ${\lambda }_2$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and 4 eV respectively then  $\frac{{\lambda }_2}{{\lambda }_1}$ is (Minimum energy to liberate an electron from ground state of hydrogen atom is 13.6 eV)	S)	$\frac{1}{2}$

 

• de Broglie wavelength of electron moving with velocity half of speed of light is equal to that of photon.

Since $\frac{1}{{\lambda }_d}\propto p$

 Then ratio $\frac{K{E}_e}{K{E}_{photon}}=\frac{{\displaystyle \frac{1}{2}}p{v}_e}{{\displaystyle \frac{1}{2}}p{v}_{photon}}=\frac{c/2}{c}=\frac{1}{2}$

•  Particle of mass M at rest decays into two particles of masses $m_1$  and  $m_2$ with non-zero velocities. Since the particles are decaying with no external force, momentum of the particles will be same. 

Since $\frac{1}{{\lambda }_d}\propto p$,  the ratio of de Broglie wavelength  $\frac{{\lambda }_1}{{\lambda }_2}=1$

• If a proton and an electron are accelerated by the same potential, then the kinetic energies are equal.  

Then the ratio,  $\frac{{\lambda }_e}{{\lambda }_p}=\frac{{\displaystyle \frac{h}{\sqrt{2m_{e}KE}}}}{{\displaystyle \frac{h}{\sqrt{2m_{p}KE}}}}=\frac{\sqrt{m_p}}{\sqrt{m_e}}>1$

• Ultraviolet light of wavelengths  ${\lambda }_1$  and  ${\lambda }_2$ when allowed to fall on hydrogen atoms in their ground state is found to liberate electrons with kinetic energy 1.8 eV and 4 eV respectively, 

Here the workfunction is 13.6 eV. 

From photoelectric equation $\frac{hc}{\lambda }=\phi +KE$

$\Rightarrow \frac{hc}{{\lambda }_1}=13.6+1.8=15.4 \mathrm{eV}$

$\Rightarrow \frac{hc}{{\lambda }_2}=13.6+4=17.6 \mathrm{eV}$

then  $\frac{{\lambda }_2}{{\lambda }_1}=\frac{15.4}{17.6}=\frac{7}{8}$