Q: The de Broglie wavelength of a neutron when its kinetic energy is \$K\$, is \$\lambda\$. What will be its wavelength when its kinetic energy is \$4K \$?

Kinetic energy of neutron=\$\dfrac{1}{2}mv^2=K\$\$\implies v=\sqrt{\dfrac{2K}{m}}\$De-broglie wavelength of neturon=\$\dfrac{h}{p}\$\$=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mK}}\propto \dfrac{1}{\sqrt{K}}\$Thus when kinetic energy becomes \$4K\$, the de-broglie wavelength becomes \$\dfrac{\lambda}{2}\$.

Q: The beam of light has three wavelengths \$4144\overset {\circ}{A}, 4972\overset {\circ}{A}\$ and \$6216\overset {\circ}{A}\$ with a total intensity of \$3.6\times 10^{-3}Wm^{2}\$ equally distributed amongst the three wavelengths. The beam falls normally on the area \$1\ cm^{2}\$ of a clean metallic surface of work function \$2.3\ eV\$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in \$2s\$.

As we know, threshold wavelength \$(\lambda_{0}) = \dfrac {hc}{\phi}\$\$\Rightarrow \lambda_{0} = \dfrac {(6.63\times 10^{-34})\times 3\times 10^{8}}{2.3\times (1.6\times 10^{-19})} = 5.404 \times 10^{-7}m\$.\$\Rightarrow \lambda_{0} = 5404\overset {\circ}{A}\$Hence, wavelength \$4144\overset {\circ}{A}\$ and \$4972\overset {\circ}{A}\$ will emit electron from the metal surface.For each wavelength energy incident on the surface per unit time\$=\$ intensity of each \$\times\$ area of the surface wavelength\$= \dfrac {3.6\times 10^{-3}}{3}\times (1\ cm)^{2} = 1.2\times 10^{-7} joule\$Therefore, energy incident on the surface for each wavelengths is \$2s\$\$E = (1.2\times 10^{-7})\times 2 = 2.4\times 10^{-7}J\$Number of photons \$n_{1}\$ due to wavelength \$4144\overset {\circ}{A}\$\$n_{1} = \dfrac {(2.4\times 10^{-7})(4144\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.5\times 10^{12}\$Number of photon \$n_{2}\$ due to the wavelength \$4972\overset {\circ}{A}\$\$n_{2} = \dfrac {(2.4\times 10^{-7})(4972\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.572\times 10^{12}\$Therefore total number of photoelectrons liberated in \$2s\$,\$N = n_{1} + n_{2}\$\$= 0.5\times 10^{12} + 0.575\times 10^{12}\$\$= 1.075\times 10^{12}\$.

Question 1

The de Broglie wavelength of a neutron when its kinetic energy is $K$, is $\lambda$. What will be its wavelength when its kinetic energy is $4K $?

Accepted Answer

Kinetic energy of neutron=$\dfrac{1}{2}mv^2=K$$\implies v=\sqrt{\dfrac{2K}{m}}$De-broglie wavelength of neturon=$\dfrac{h}{p}$$=\dfrac{h}{mv}=\dfrac{h}{\sqrt{2mK}}\propto \dfrac{1}{\sqrt{K}}$Thus when kinetic energy becomes $4K$, the de-broglie wavelength becomes $\dfrac{\lambda}{2}$.

Question 2

The beam of light has three wavelengths $4144\overset {\circ}{A}, 4972\overset {\circ}{A}$ and $6216\overset {\circ}{A}$ with a total intensity of $3.6\times 10^{-3}Wm^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1\ cm^{2}$ of a clean metallic surface of work function $2.3\ eV$. Assume that there is no loss of light by reflection and that each energetically capable photon ejects one electron. Calculate the number of photoelectrons liberated in $2s$.

Accepted Answer

As we know, threshold wavelength $(\lambda_{0}) = \dfrac {hc}{\phi}$$\Rightarrow \lambda_{0} = \dfrac {(6.63\times 10^{-34})\times 3\times 10^{8}}{2.3\times (1.6\times 10^{-19})} = 5.404 \times 10^{-7}m$.$\Rightarrow \lambda_{0} = 5404\overset {\circ}{A}$Hence, wavelength $4144\overset {\circ}{A}$ and $4972\overset {\circ}{A}$ will emit electron from the metal surface.For each wavelength energy incident on the surface per unit time$=$ intensity of each $\times$ area of the surface wavelength$= \dfrac {3.6\times 10^{-3}}{3}\times (1\ cm)^{2} = 1.2\times 10^{-7} joule$Therefore, energy incident on the surface for each wavelengths is $2s$$E = (1.2\times 10^{-7})\times 2 = 2.4\times 10^{-7}J$Number of photons $n_{1}$ due to wavelength $4144\overset {\circ}{A}$$n_{1} = \dfrac {(2.4\times 10^{-7})(4144\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.5\times 10^{12}$Number of photon $n_{2}$ due to the wavelength $4972\overset {\circ}{A}$$n_{2} = \dfrac {(2.4\times 10^{-7})(4972\times 10^{-10})}{(6.63\times 10^{-34})(3\times 10^{8})} = 0.572\times 10^{12}$Therefore total number of photoelectrons liberated in $2s$,$N = n_{1} + n_{2}$$= 0.5\times 10^{12} + 0.575\times 10^{12}$$= 1.075\times 10^{12}$.

Dual Nature Of Radiation And Matter

Physics
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The de Broglie wavelength of a neutron when its kinetic energy is $K$ , is $λ$ . What will be its wavelength when its kinetic energy is $4 K$ ?

Dual Nature Of Radiation And Matter

Physics 1009 Views

The de Broglie wavelength of a neutron when its kinetic energy is K, is λ. What will be its wavelength when its kinetic energy is 4K?

Physics
1009 Views

The de Broglie wavelength of a neutron when its kinetic energy is $K$ , is $λ$ . What will be its wavelength when its kinetic energy is $4 K$ ?