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}\$

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 (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: The wavelength \$\lambda _{e}\$ of an electron and \$\lambda _{p}\$ of a photon of same energy E are related by :

Wavelength of electron, \$\displaystyle \lambda _{e}=\dfrac{h}{\sqrt{2mE}}\$ Wavelength of photon, \$\displaystyle \lambda _{p}=\dfrac{hc}{E}\$\$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2mE}\$\$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2m\dfrac{hc}{\lambda _{p}}}\\ \Rightarrow \lambda _{e}^{2}\propto \lambda _{p}\$

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: What is the De-Broglie wavelength associated with the hydrogen electron in its third orbit :

Radius of 3rd orbit \$r=R\times 3^2=9R\$Since,\$\dfrac{nh}{2\pi}=mvr\$\$\dfrac{h}{mv}=\dfrac{2\pi r}{n}\$\$\lambda=\dfrac{h}{mv}=\dfrac{2\pi \times 9R}{3}\$\$\lambda=6\pi R\$We know that\$R=0.529 A^0\$Therefore,\$\lambda=6\pi \times 0.529\$\$\lambda=9.96 A^0\$\$\lambda=9.96 \times 10^{-8}\ cm\$Hence, this is the answer.

Q: If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of an electron will be:

According to de-Broglie relation:\$\lambda = \dfrac{h}{p}\$Let, initial momentum \$=p_1\$ and associated wavelength is \$\lambda_1\$ and with a decrease in wavelength, momentum will increase.Thus, \$\Delta p=-P\$ and for \$\Delta \lambda=\dfrac {0.5}{100}\times \lambda_1\$Differentiating the de-Broglie relation we get:\$\dfrac { \Delta \lambda }{\lambda_1}=-\dfrac{ \Delta p }{p_1}\$and \$\dfrac { 0.005\times \lambda_1 }{\lambda_1}=\dfrac{ P }{p_1}=\dfrac{0.5}{100}\$\$p_1=200P\$Option D is correct.

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 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 3

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

Question 4

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 5

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 6

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 7

The wavelength $\lambda _{e}$ of an electron and $\lambda _{p}$ of a photon of same energy E are related by :

Accepted Answer

Wavelength of electron, $\displaystyle \lambda _{e}=\dfrac{h}{\sqrt{2mE}}$  Wavelength of photon, $\displaystyle \lambda _{p}=\dfrac{hc}{E}$$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2mE}$$\displaystyle \lambda _{e}^{2}=\dfrac{h^{2}}{2m\dfrac{hc}{\lambda _{p}}}\ \Rightarrow  \lambda _{e}^{2}\propto \lambda _{p}$

Question 8

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:

Accepted Answer

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, $\lambda\geq 2.8\times 10^{-9}  m$

Question 9

What is the De-Broglie wavelength associated with the hydrogen electron in its third orbit :

Accepted Answer

Radius of 3rd orbit $r=R\times 3^2=9R$Since,$\dfrac{nh}{2\pi}=mvr$$\dfrac{h}{mv}=\dfrac{2\pi r}{n}$$\lambda=\dfrac{h}{mv}=\dfrac{2\pi \times 9R}{3}$$\lambda=6\pi R$We know that$R=0.529 A^0$Therefore,$\lambda=6\pi \times 0.529$$\lambda=9.96 A^0$$\lambda=9.96 \times 10^{-8}\ cm$Hence, this is the answer.

Question 10

If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of an electron will be:

Accepted Answer

According to de-Broglie relation:$\lambda = \dfrac{h}{p}$Let, initial momentum $=p_1$ and associated wavelength is $\lambda_1$ and with a decrease in wavelength, momentum will increase.Thus, $\Delta p=-P$ and for $\Delta \lambda=\dfrac {0.5}{100}\times \lambda_1$Differentiating the de-Broglie relation we get:$\dfrac { \Delta \lambda }{\lambda_1}=-\dfrac{ \Delta p }{p_1}$and $\dfrac { 0.005\times \lambda_1 }{\lambda_1}=\dfrac{ P }{p_1}=\dfrac{0.5}{100}$$p_1=200P$Option D is correct.

Quantum Mechanics.

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REVISE WITH CONCEPTS

Wave Nature of Matter

QUICK SUMMARY WITH STORIES

De Broglie Wavelength

Heisenberg Uncertainty Principle

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

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

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

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 :

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)

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 :

The wavelength $λ_{e}$ of an electron and $λ_{p}$ of a photon of same energy E are related by :

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is:

What is the De-Broglie wavelength associated with the hydrogen electron in its third orbit :

If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of an electron will be:

Quantum Mechanics.

Language

Rate

REVISE WITH CONCEPTS

Wave Nature of Matter

QUICK SUMMARY WITH STORIES

De Broglie Wavelength

Heisenberg Uncertainty Principle

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

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

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

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 :

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)

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 :

The wavelength λe​ of an electron and λp​ of a photon of same energy E are related by :

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:

What is the De-Broglie wavelength associated with the hydrogen electron in its third orbit :

If the momentum of an electron is changed by P, then the de-Broglie wavelength associated with it changes by 0.5%. The initial momentum of an electron will be:

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 :

The wavelength $λ_{e}$ of an electron and $λ_{p}$ of a photon of same energy E are related by :

Light of wavelength $500 n m$ is incident on a metal with a work function $2.28 e V$ . The de Borglie wavelength of the emitted electron is: