Hard X-rays for the study of fractures in bones should have a minimum wavelength of \displaystyle { 10 }^{ -11 }m . The accelerating voltage for electrons in X-ray machine should be:

From conservation of energy the electron kinetic energy equals the maximum photon energy (we neglect the work function \displaystyle \phi because it is normally so small compared to \displaystyle { eV }_{ 0 } \displaystyle { eV }_{ 0 }={ hv }_{ max } \displaystyle { eV }_{ 0 }=\frac { hc }{ { \lambda }_{ min } } \displaystyle { V }_{ 0 }=\frac { hc }{ { e\lambda }_{ min } } h (Plank's Constant) = 6.625\times 10^{-34}J-s , c (Speed of light) = 3\times 10^{8}\ m/s , e (Charge of electron) = 1.6\times10^{-19}C \displaystyle { V }_{ 0 }=\frac { 12400\times { 10 }^{ -10 } }{ { 10 }^{ -11 } } \displaystyle =124\ kV Hence, accelerating voltage for electrons in X-ray machine should be less than 124\ kV .

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} .

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Dual Nature Of Radiation And Matter

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} .

Physics
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JEE Mains JEE Advanced NEET AIIMS BITSAT Easy Qs Med Qs Hard Qs

Hard X-rays for the study of fractures in bones should have a minimum wavelength of $10^{- 11} m$ . The accelerating voltage for electrons in X-ray machine should be:

< 124 k V

> 124 k V

Between

60 k V

and

70 k V

= 100 k V

Medium

BITSAT

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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 $4 K$ ?

\frac{λ}{4}

\frac{λ}{2}

2 λ

4 λ

Medium

BITSAT

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The beam of light has three wavelengths $4144 A \circ, 4972 A \circ$ and $6216 A \circ$ with a total intensity of $3.6 \times 1 0^{- 3} W m^{2}$ equally distributed amongst the three wavelengths. The beam falls normally on the area $1 c m^{2}$ of a clean metallic surface of work function $2.3 e V$ . 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 $2 s$ .

2 \times 1 0^{9}

1.075 \times 1 0^{12}

9 \times 1 0^{8}

3.75 \times 1 0^{6}

Hard

BITSAT

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Dual Nature Of Radiation And Matter

Physics 151 Views

Hard X-rays for the study of fractures in bones should have a minimum wavelength of 10−11m. The accelerating voltage for electrons in X-ray machine should be:

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
151 Views

Hard X-rays for the study of fractures in bones should have a minimum wavelength of $10^{- 11} m$ . The accelerating voltage for electrons in X-ray machine should be:

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 $4 K$ ?