(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.Given values are $h = 6.63 \times 10^{-34} \ J \cdot s$,$m_{

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(B) The de-Broglie wavelength $\lambda$ is given by the formula: $\lambda = \frac{h}{mv}$.Given values are $h = 6.63 \times 10^{-34} \ J \cdot s$,$m_{e} = 9.11 \times 10^{-31} \ kg$,and $v = 6.4 \times 10^{6} \ m/s$.Substituting these values into the formula:$\lambda = \frac{6.63 \times 10^{-34}}{9.11 \times 10^{-31} \times 6.4 \times 10^{6}}$$\lambda = \frac{6.63 \times 10^{-34}}{58.304 \times 10^{-25}}$$\lambda \approx 0.1137 \times 10^{-9} \ m$Since $1 \ nm = 10^{-9} \ m$,we get $\lambda \approx 0.114 \ nm$.

Class 12 Physics Dual Nature of Radiation and matter Matter Waves and de Broglie Wavelength

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What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

GUJCET 2022 All GUJCET

A
$0.124$
B
$0.114$
C
$0.135$
D
$0.145$

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Dual Nature of Radiation and matter

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Matter Waves and de Broglie Wavelength

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$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
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The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .

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What is the de-Broglie wavelength associated with an electron moving with a speed of $6.4 \times 10^{6} \ m/s$ (in $nm$)? $(m_{e} = 9.11 \times 10^{-31} \ kg, h = 6.63 \times 10^{-34} \ J \cdot s)$

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$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)

If a photon,an electron,and a uranium nucleus have the same de Broglie wavelength,which one will have the highest energy?

The energy that should be added to an electron,to reduce its de-Broglie wavelength from $10^{-10} \ m$ to $0.5 \times 10^{-10} \ m$,will be

The ratio of the de-Broglie wavelength of molecules of hydrogen $(H_2)$ and helium $(He)$ which are at temperatures $27^{\circ} C$ and $127^{\circ} C$ respectively is:

The relation between the wavelength of electromagnetic radiation $(\lambda)$ and the de Broglie wavelength of its quantum (photon) $(\lambda')$ is . . . . . . .

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$A$ proton and an electron are associated with the same de-Broglie wavelength. The ratio of their kinetic energies is:
(Assume $h=6.63 \times 10^{-34} \ J \ s$,$m_{e}=9.0 \times 10^{-31} \ kg$ and $m_{p}=1836 \times m_{e}$)