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Wave Nature of Matter - Problem L1

Q: A proton and an alpha particle both are accelerated through the same potential difference. The ratio of corresponding de-Broglie wavelengths is :

Key Point : The de-Broglie wavelength of a particle of mass m and moving with velocity \$v\$ is given by\$\lambda = \dfrac {h}{mv} \$ \$ (\because p = mv)\$de-Broglie wavelength of a proton of mass \$m_{1}\$ and kinetic energy \$k\$ is given by\$\lambda_{1} = \dfrac {h}{\sqrt {2m_{1}k}}\$ \$(\because p = \sqrt {2mk})\$\$\lambda_1= \dfrac {h}{\sqrt {2m_{1}qV}} .... (i)\$ \$ [\because k = qV]\$For an alpha particle mass \$m_{2}\$ carrying charge \$q_{0}\$ is accelerated through potential \$V\$, then\$\lambda_{2} = \dfrac {h}{\sqrt {2m_{2}q_{0}V}}\$\$\because\$ For \$\alpha - particle\ \ (^{4}_{2}He)\$ : \$ q_{0} = 2q\$ and \$m_{2} = 4m_{1}\$\$\therefore \lambda_{2} = \dfrac {h}{\sqrt {2\times 4m_{1}\times 2q\times V}} .... (ii)\$The ratio of corresponding wavelength, from Eqs. (i) and (ii), we get\$\dfrac {\lambda_{1}}{\lambda_{2}} = \dfrac {h}{\sqrt {2m_{1}qV}} \times \dfrac {\sqrt {2\times m_{1} \times 4\times 2qV}}{h} = \dfrac {4}{\sqrt {2}} \times \dfrac {\sqrt {2}}{\sqrt {2}}\$We get \$\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}\$

Q: An electron (of mass m) and a photon have the same energy E in the range of a few eV. The ratio of the de-Broglie wavelength associated with the electron and the wavelength of the photon is (c = speed of light in vacuum)

For electronDe - Broglie wavelength \$\lambda_c = \dfrac{h}{p}\$where p is momentum \$p = mv\$also by energy we have \$E = \dfrac{1}{2} mv^2\$\$\Rightarrow E = \dfrac{1}{2} \dfrac{p^2}{m}\$\$\Rightarrow p = \sqrt{2 mE}\$\$\therefore \lambda_c = \dfrac{h}{\sqrt{2 mE}}\$For photon energy \$\Rightarrow E = \dfrac{hc}{\lambda}\$\$\Rightarrow \lambda = \dfrac{h c}{E}\$\$\therefore \dfrac{\lambda_c}{\lambda} = \dfrac{h}{\sqrt{2 mE}} \dfrac{E}{hc}\$\$= \dfrac{1}{C} \sqrt{\dfrac{E}{2m}}\$option (A) is correct.

Q: An electromagnetic wave of wavelength \$\lambda\$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength \$\lambda'\$, then

Kinetic energy of emitted electron = Energy of incident photoni.e. \$\dfrac{1}{2}mv^2 = h \nu\$or \$\dfrac{p^2}{2m} = \dfrac{hc}{\lambda}\$\$\left[ \because mv = p, \nu =\dfrac{c}{\lambda}\right]\$or \$p = \sqrt{\dfrac{2mhc}{\lambda}}\$de-Broglie wavelength of emitted electrons \$\lambda' = \dfrac{h}{p}=\dfrac{h}{\sqrt{\dfrac{2mhc}{\lambda}}}\$ or \$\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}\$ \$\therefore \lambda = \dfrac{2mc}{h}\lambda'^2\$

Q: An electron is accelerated from rest through a potential difference of \$V\$ volt. If the de Broglie wavelength of the electron is \$1.227 \times 10^{-2} nm\$, the potential difference is :

The de-broglie wavelength of an electron is given as:\$\lambda=\dfrac{1.227}{\sqrt{V}}\ nm\$Substitute the wavelength in the above expression:\$V=\left(\dfrac{1.227}{1.227\times10^{-2}}\right)^2\$\$V=10^4\ V\$

Q: An electron, an alpha particle and a proton have the same kinetic energy. Which one of these particles has (i) the shortest and (ii) the largest, de Broglie wavelength?

De Broglie wavelength of particle of mass m in terms of kinetic energy \$(E_k)\$ is \$\lambda=\dfrac{h}{\sqrt{2mE_k}} \propto \dfrac{1}{m}\$ for the same kinetic energy.(i) Out of given particles, the mass of alpha particle is maximum, so de Broglie wavelength associated with alpha particle is the shortest.(ii) As mass of electron is least, so electron has largest de-Broglie wavelength.

Q: An \$\alpha\$- particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

Given that;\$V_{\alpha}=V_{Pr}\$Since:- \$\lambda=\dfrac{h}{p}\$\$\Rightarrow \lambda=\dfrac{h}{\sqrt{2mE}}\$Where \$E=kinetic\ Energy=eV\$ [V=acceleration potential]\$\Rightarrow \lambda_{Pr}=\dfrac{h}{\sqrt{2m_p(2e)V_{Pr}}}\$Whereas; \$\lambda_{\alpha}=\dfrac{h}{\sqrt{m_{\alpha}(2e)V_{\alpha}}}\$Where \$m_{\alpha}=4m_{Pr}\$and \$V_{\alpha}=V_{Pr}=V\$\$\Rightarrow \lambda_{Pr}=\dfrac{h}{\sqrt{2m_p(e)V}}\$...........(1)and \$\lambda_{\alpha}=\dfrac{h}{\sqrt{2(4mp)(2e)V}}\$...........(2)From (1) and (2)\$\dfrac{\lambda_{Pr}}{\lambda_{\alpha}}=\dfrac{h}{\sqrt{2m_p(e)V}}\times \dfrac{\sqrt{2(4m_p)(2e)V}}{h}\$\$=\dfrac{\sqrt{2m_peV(8)}}{\sqrt{2m_peV}}\$\$=\sqrt{8}:1\$

Q: Light of wavelength \$500\ nm\$ is incident on a metal with a work function \$2.28\ eV\$. The de Borglie wavelength of the emitted electron is:

From photoelectric effect, \$h\nu=W_0+eV\$or \$\dfrac{hc}{\lambda}=W_0+eV\$or \$\displaystyle \frac{(6.6\times 10^{-34})\times (3\times 10^8)}{500\times 10^{-9}}=2.28\times 1.6\times 10^{-19}+eV\$or \$eV=0.31 \times 10^{-19}\$ or \$V=0.2\$Here V should be \$V\leq 0.2\$The de Broglie wavelength of electron , \$\lambda=\dfrac{12.27\times 10^{-10}}{V}\$so, \$\lambda\geq 2.8\times 10^{-9} m\$

Q: Derive an expression for de Broglie wavelength of matter waves.

de Broglie equated the energy equations of Planck (wave) and Einstein (particle).\$E=hv\$ (planck's relation)\$E=m{ c }^{ 2 }\$ (Einstein's mass - energy relation)\$hv=m{ c }^{ 2 }\$\$\dfrac { hc }{ \lambda } =m{ c }^{ 2 }\$ (or)\$\lambda =\dfrac { h }{ mc }\$For a particle moving with a velocity \$V\$,\$\lambda =\dfrac { h }{ mV } =\dfrac { h }{ p }\$where \$p=mV\$ is the momentum of the particle.

Q: What is the de Broglie wavelength of the electron accelerated through a potential difference of 100 Volt?

Given : \$V =100\$ voltsde-Broglie wavelength of electron \$\lambda = \dfrac{h}{\sqrt{2meV}}\$\$\therefore\$ \$\lambda = \dfrac{6.6\times 10^{-34}}{\sqrt{2(9.1\times 10^{-31})(1.6\times 10^{-19})(100)}}\$Or \$\lambda = \dfrac{6.6\times 10^{-34}}{5.4\times 10^{-10}}\$ m\$\implies \$ \$\lambda = 1.227\times 10^{-10} m = 1.227\$ \$\mathring{A}\$

Physics

definition

de Broglie wavelength

If accelerating potential V is applied ti an electron beam, the energy acquired by the electron is expressed in electron-volt which is equal to charge on the electron in coulombs x potential in volts. This energy becomes the K.E of the electron.

λ = \frac{h}{m v} = \frac{h}{( 2 m e V )}

λ = \frac{1 . 2 2 6 \times 1 0 ^{- 9}}{V}

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