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 \$\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: 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: 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}\$

Question 1

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

Accepted Answer

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  $\dfrac{\lambda_1}{\lambda_2}= 2\sqrt {2}$

Question 2

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)

Accepted Answer

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.

Question 3

Derive an expression for de Broglie wavelength of matter waves.

Accepted Answer

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.

Question 4

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?

Accepted Answer

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.

Question 5

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

Accepted Answer

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$

Question 6

Light with an energy flux of $18 w/cm^2$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 cm^2$. Find the average force exerted on the surface during a 30 minute time span.

Accepted Answer

Question 7

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

Accepted Answer

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 $\lambda' = \sqrt{\dfrac{h\lambda}{2mc}}$ $\therefore \lambda = \dfrac{2mc}{h}\lambda'^2$

Question 8

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

Accepted Answer

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Question 9

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 :

Accepted Answer

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$

Question 10

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

Accepted Answer

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 $   $\lambda = 1.227\times 10^{-10} m = 1.227$  $\mathring{A}$

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Wave Nature of Matter - Problem L1

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De Broglie Wavelength

Heisenberg Uncertainty Principle

Wave Nature of Matter - Problem L1

Wave Nature of Matter - Problem L1

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

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)

Derive an expression for de Broglie wavelength of matter waves.

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?

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

Light with an energy flux of $18 w / c m^{2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 c m^{2}$ . Find the average force exerted on the surface during a 30 minute time span.

An electromagnetic wave of wavelength $λ$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $λ^{'}$ , then

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 1 0^{- 2} n m$ , the potential difference is :

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

Revise with Concepts

Wave Nature of Matter

Wave Nature of Matter - Problem L1

Learn with Videos

Quick Summary With Stories

De Broglie Wavelength

Heisenberg Uncertainty Principle

Wave Nature of Matter - Problem L1

Wave Nature of Matter - Problem L1

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

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)

Derive an expression for de Broglie wavelength of matter waves.

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?

An α- particle and a proton are accelerated from rest by the same potential. Find the ratio of their de- Broglie wavelength.

Light with an energy flux of 18w/cm2 falls on a non-reflecting surface at normal incidence. If the surface has an area of 20cm2. Find the average force exerted on the surface during a 30 minute time span.

An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength λ′, then

If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is :

An electron is accelerated from rest through a potential difference of V volt. If the de Broglie wavelength of the electron is 1.227×10−2nm, the potential difference is :

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

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

Light with an energy flux of $18 w / c m^{2}$ falls on a non-reflecting surface at normal incidence. If the surface has an area of $20 c m^{2}$ . Find the average force exerted on the surface during a 30 minute time span.

An electromagnetic wave of wavelength $λ$ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Brogue wavelength $λ^{'}$ , then

An electron is accelerated from rest through a potential difference of $V$ volt. If the de Broglie wavelength of the electron is $1.227 \times 1 0^{- 2} n m$ , the potential difference is :