142519
Find the de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
1 $112 \mathrm{pm}$
2 $95 \mathrm{pm}$
3 $124 \mathrm{pm}$
4 $102 \mathrm{pm}$
Explanation:
A Given, Kinetic energy electron $=120 \mathrm{eV}$ $=120 \times 1.6 \times 10^{-19} \mathrm{~J}$ For an electron $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}}$ $\lambda=112 \times 10^{-12} \mathrm{~m}$ $\lambda=112 \mathrm{pm}$ $\left[\therefore 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]$
Karnataka CET-2015
Dual nature of radiation and Matter
142520
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}$ is :
1 1
2 1836
3 $\frac{1}{1836}$
4 918
Explanation:
B We know, mass of proton $\left(\mathrm{m}_{\mathrm{p}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ and mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{v}_{\mathrm{p}}=\mathrm{v}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}}$ $(\because \text { speed is same) }$ From equation II/I $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{p}} \mathrm{v}}{\mathrm{h}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=1.8351 \times 10^{3}$ $=1835.1$ $\square 1836$
Karnataka CET-2011
Dual nature of radiation and Matter
142521
The de Broglie wavelength of an electron in the first Bohr orbit is :
1 equal to half the circumference of the first orbit
2 equal to one fourth the circumference of the first orbit
3 equal to the circumference of the first orbit
4 equal to twice the circumference of the first orbit
Explanation:
C We know, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\operatorname{mvr}=\frac{\mathrm{h}}{2 \pi}$ From equation (i) and (ii) $\lambda=2 \pi \mathrm{r}$ Hence, the equal to the circumference of first orbit.
Karnataka CET-2002
Dual nature of radiation and Matter
142522
The de Broglie wavelength $\lambda$ of an electron accelerated through a potential $\mathrm{V}$ (in volts) is
1 $\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
2 $\frac{0.1227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
3 $\frac{0.01227}{\sqrt{V}} \mathrm{~nm}$
4 $\frac{0.1227}{\sqrt{\mathrm{V}}} \AA$
Explanation:
A We know that, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \mathrm{V}}}$ $\lambda=\frac{1.227 \times 10^{-9}}{\sqrt{\mathrm{V}}}$ $\lambda=\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
142519
Find the de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
1 $112 \mathrm{pm}$
2 $95 \mathrm{pm}$
3 $124 \mathrm{pm}$
4 $102 \mathrm{pm}$
Explanation:
A Given, Kinetic energy electron $=120 \mathrm{eV}$ $=120 \times 1.6 \times 10^{-19} \mathrm{~J}$ For an electron $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}}$ $\lambda=112 \times 10^{-12} \mathrm{~m}$ $\lambda=112 \mathrm{pm}$ $\left[\therefore 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]$
Karnataka CET-2015
Dual nature of radiation and Matter
142520
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}$ is :
1 1
2 1836
3 $\frac{1}{1836}$
4 918
Explanation:
B We know, mass of proton $\left(\mathrm{m}_{\mathrm{p}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ and mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{v}_{\mathrm{p}}=\mathrm{v}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}}$ $(\because \text { speed is same) }$ From equation II/I $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{p}} \mathrm{v}}{\mathrm{h}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=1.8351 \times 10^{3}$ $=1835.1$ $\square 1836$
Karnataka CET-2011
Dual nature of radiation and Matter
142521
The de Broglie wavelength of an electron in the first Bohr orbit is :
1 equal to half the circumference of the first orbit
2 equal to one fourth the circumference of the first orbit
3 equal to the circumference of the first orbit
4 equal to twice the circumference of the first orbit
Explanation:
C We know, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\operatorname{mvr}=\frac{\mathrm{h}}{2 \pi}$ From equation (i) and (ii) $\lambda=2 \pi \mathrm{r}$ Hence, the equal to the circumference of first orbit.
Karnataka CET-2002
Dual nature of radiation and Matter
142522
The de Broglie wavelength $\lambda$ of an electron accelerated through a potential $\mathrm{V}$ (in volts) is
1 $\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
2 $\frac{0.1227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
3 $\frac{0.01227}{\sqrt{V}} \mathrm{~nm}$
4 $\frac{0.1227}{\sqrt{\mathrm{V}}} \AA$
Explanation:
A We know that, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \mathrm{V}}}$ $\lambda=\frac{1.227 \times 10^{-9}}{\sqrt{\mathrm{V}}}$ $\lambda=\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
142519
Find the de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
1 $112 \mathrm{pm}$
2 $95 \mathrm{pm}$
3 $124 \mathrm{pm}$
4 $102 \mathrm{pm}$
Explanation:
A Given, Kinetic energy electron $=120 \mathrm{eV}$ $=120 \times 1.6 \times 10^{-19} \mathrm{~J}$ For an electron $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}}$ $\lambda=112 \times 10^{-12} \mathrm{~m}$ $\lambda=112 \mathrm{pm}$ $\left[\therefore 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]$
Karnataka CET-2015
Dual nature of radiation and Matter
142520
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}$ is :
1 1
2 1836
3 $\frac{1}{1836}$
4 918
Explanation:
B We know, mass of proton $\left(\mathrm{m}_{\mathrm{p}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ and mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{v}_{\mathrm{p}}=\mathrm{v}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}}$ $(\because \text { speed is same) }$ From equation II/I $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{p}} \mathrm{v}}{\mathrm{h}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=1.8351 \times 10^{3}$ $=1835.1$ $\square 1836$
Karnataka CET-2011
Dual nature of radiation and Matter
142521
The de Broglie wavelength of an electron in the first Bohr orbit is :
1 equal to half the circumference of the first orbit
2 equal to one fourth the circumference of the first orbit
3 equal to the circumference of the first orbit
4 equal to twice the circumference of the first orbit
Explanation:
C We know, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\operatorname{mvr}=\frac{\mathrm{h}}{2 \pi}$ From equation (i) and (ii) $\lambda=2 \pi \mathrm{r}$ Hence, the equal to the circumference of first orbit.
Karnataka CET-2002
Dual nature of radiation and Matter
142522
The de Broglie wavelength $\lambda$ of an electron accelerated through a potential $\mathrm{V}$ (in volts) is
1 $\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
2 $\frac{0.1227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
3 $\frac{0.01227}{\sqrt{V}} \mathrm{~nm}$
4 $\frac{0.1227}{\sqrt{\mathrm{V}}} \AA$
Explanation:
A We know that, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \mathrm{V}}}$ $\lambda=\frac{1.227 \times 10^{-9}}{\sqrt{\mathrm{V}}}$ $\lambda=\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
142519
Find the de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
1 $112 \mathrm{pm}$
2 $95 \mathrm{pm}$
3 $124 \mathrm{pm}$
4 $102 \mathrm{pm}$
Explanation:
A Given, Kinetic energy electron $=120 \mathrm{eV}$ $=120 \times 1.6 \times 10^{-19} \mathrm{~J}$ For an electron $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}}$ $\lambda=112 \times 10^{-12} \mathrm{~m}$ $\lambda=112 \mathrm{pm}$ $\left[\therefore 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]$
Karnataka CET-2015
Dual nature of radiation and Matter
142520
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}$ is :
1 1
2 1836
3 $\frac{1}{1836}$
4 918
Explanation:
B We know, mass of proton $\left(\mathrm{m}_{\mathrm{p}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ and mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{v}_{\mathrm{p}}=\mathrm{v}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}}$ $(\because \text { speed is same) }$ From equation II/I $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{p}} \mathrm{v}}{\mathrm{h}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=1.8351 \times 10^{3}$ $=1835.1$ $\square 1836$
Karnataka CET-2011
Dual nature of radiation and Matter
142521
The de Broglie wavelength of an electron in the first Bohr orbit is :
1 equal to half the circumference of the first orbit
2 equal to one fourth the circumference of the first orbit
3 equal to the circumference of the first orbit
4 equal to twice the circumference of the first orbit
Explanation:
C We know, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\operatorname{mvr}=\frac{\mathrm{h}}{2 \pi}$ From equation (i) and (ii) $\lambda=2 \pi \mathrm{r}$ Hence, the equal to the circumference of first orbit.
Karnataka CET-2002
Dual nature of radiation and Matter
142522
The de Broglie wavelength $\lambda$ of an electron accelerated through a potential $\mathrm{V}$ (in volts) is
1 $\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
2 $\frac{0.1227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
3 $\frac{0.01227}{\sqrt{V}} \mathrm{~nm}$
4 $\frac{0.1227}{\sqrt{\mathrm{V}}} \AA$
Explanation:
A We know that, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \mathrm{V}}}$ $\lambda=\frac{1.227 \times 10^{-9}}{\sqrt{\mathrm{V}}}$ $\lambda=\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
142519
Find the de-Broglie wavelength of an electron with kinetic energy of $120 \mathrm{eV}$.
1 $112 \mathrm{pm}$
2 $95 \mathrm{pm}$
3 $124 \mathrm{pm}$
4 $102 \mathrm{pm}$
Explanation:
A Given, Kinetic energy electron $=120 \mathrm{eV}$ $=120 \times 1.6 \times 10^{-19} \mathrm{~J}$ For an electron $(\lambda)=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}_{\mathrm{e}} \mathrm{KE}_{\mathrm{e}}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 120 \times 1.6 \times 10^{-19}}}$ $\lambda=112 \times 10^{-12} \mathrm{~m}$ $\lambda=112 \mathrm{pm}$ $\left[\therefore 1 \mathrm{pm}=10^{-12} \mathrm{~m}\right]$
Karnataka CET-2015
Dual nature of radiation and Matter
142520
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are moving with the same speed. The ratio of their de-Broglie's wavelengths $\lambda_{\mathrm{e}} / \lambda_{\mathrm{p}}$ is :
1 1
2 1836
3 $\frac{1}{1836}$
4 918
Explanation:
B We know, mass of proton $\left(\mathrm{m}_{\mathrm{p}}\right)=1.67 \times 10^{-27} \mathrm{~kg}$ and mass of electron $\left(\mathrm{m}_{\mathrm{e}}\right)=9.1 \times 10^{-31} \mathrm{~kg}$ de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ $\mathrm{v}_{\mathrm{e}}=\mathrm{v}_{\mathrm{p}}=\mathrm{v}$ $\lambda_{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{p}} \mathrm{v}}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}_{\mathrm{e}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}}$ $(\because \text { speed is same) }$ From equation II/I $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{h}}{\mathrm{m}_{\mathrm{e}} \mathrm{v}} \times \frac{\mathrm{m}_{\mathrm{p}} \mathrm{v}}{\mathrm{h}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}=\frac{1.67 \times 10^{-27}}{9.1 \times 10^{-31}}$ $=1.8351 \times 10^{3}$ $=1835.1$ $\square 1836$
Karnataka CET-2011
Dual nature of radiation and Matter
142521
The de Broglie wavelength of an electron in the first Bohr orbit is :
1 equal to half the circumference of the first orbit
2 equal to one fourth the circumference of the first orbit
3 equal to the circumference of the first orbit
4 equal to twice the circumference of the first orbit
Explanation:
C We know, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}} \quad(\therefore \mathrm{p}=\mathrm{mv})$ $\operatorname{mvr}=\frac{\mathrm{h}}{2 \pi}$ From equation (i) and (ii) $\lambda=2 \pi \mathrm{r}$ Hence, the equal to the circumference of first orbit.
Karnataka CET-2002
Dual nature of radiation and Matter
142522
The de Broglie wavelength $\lambda$ of an electron accelerated through a potential $\mathrm{V}$ (in volts) is
1 $\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
2 $\frac{0.1227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
3 $\frac{0.01227}{\sqrt{V}} \mathrm{~nm}$
4 $\frac{0.1227}{\sqrt{\mathrm{V}}} \AA$
Explanation:
A We know that, de Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mk}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\lambda=\frac{6.63 \times 10^{-34}}{\sqrt{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times \mathrm{V}}}$ $\lambda=\frac{1.227 \times 10^{-9}}{\sqrt{\mathrm{V}}}$ $\lambda=\frac{1.227}{\sqrt{\mathrm{V}}} \mathrm{nm}$
