142500
The de-Broglie wavelength of $1 \mathrm{~kg}$ mass moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$, will be :
1 $6.626 \times 10^{-35} \mathrm{~m}$
2 $6.626 \times 10^{-34} \mathrm{~m}$
3 $6.626 \times 10^{-33} \mathrm{~m}$
4 none of these
Explanation:
A Given that, $\mathrm{m}=1 \mathrm{~kg}, \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\lambda=\frac{6.626 \times 10^{-34}}{1 \times 10}$ $\lambda=6.63 \times 10^{-35} \mathrm{~m}$
JCECE-2004
Dual nature of radiation and Matter
142502
If the momentum of a particle is doubled, then its de-Broglie wavelength will :
1 become four times
2 become half
3 become two times
4 remain unchanged
Explanation:
B From de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\mathrm{p}=\mathrm{mv})$ Hence if the momentum of a particle is doubled, value of $\lambda$ will be halved.
JCECE-2003
Dual nature of radiation and Matter
142509
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of $1 \mathrm{kV}$, its kinetic energy will be
1 $1840 \mathrm{keV}$
2 $1 / 1840 \mathrm{keV}$
3 $1 \mathrm{keV}$
4 $920 \mathrm{~V}$
Explanation:
C Given that, Mass of proton $=1840 \times$ mass of an electron Final kinetic energy - Initial kinetic $=$ Total change in energy. K.E $-0=$ Charge $\times$ potential difference $\mathrm{K} . \mathrm{E}=\mathrm{e} \times 1 \mathrm{kV}=1 \mathrm{keV}$
AIIMS-2003
Dual nature of radiation and Matter
142510
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B As the electron and photon are having the same wavelength $\lambda$, the momentum of both of them will be the same. According to de-Broglie wavelength of a particle or photon, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
142500
The de-Broglie wavelength of $1 \mathrm{~kg}$ mass moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$, will be :
1 $6.626 \times 10^{-35} \mathrm{~m}$
2 $6.626 \times 10^{-34} \mathrm{~m}$
3 $6.626 \times 10^{-33} \mathrm{~m}$
4 none of these
Explanation:
A Given that, $\mathrm{m}=1 \mathrm{~kg}, \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\lambda=\frac{6.626 \times 10^{-34}}{1 \times 10}$ $\lambda=6.63 \times 10^{-35} \mathrm{~m}$
JCECE-2004
Dual nature of radiation and Matter
142502
If the momentum of a particle is doubled, then its de-Broglie wavelength will :
1 become four times
2 become half
3 become two times
4 remain unchanged
Explanation:
B From de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\mathrm{p}=\mathrm{mv})$ Hence if the momentum of a particle is doubled, value of $\lambda$ will be halved.
JCECE-2003
Dual nature of radiation and Matter
142509
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of $1 \mathrm{kV}$, its kinetic energy will be
1 $1840 \mathrm{keV}$
2 $1 / 1840 \mathrm{keV}$
3 $1 \mathrm{keV}$
4 $920 \mathrm{~V}$
Explanation:
C Given that, Mass of proton $=1840 \times$ mass of an electron Final kinetic energy - Initial kinetic $=$ Total change in energy. K.E $-0=$ Charge $\times$ potential difference $\mathrm{K} . \mathrm{E}=\mathrm{e} \times 1 \mathrm{kV}=1 \mathrm{keV}$
AIIMS-2003
Dual nature of radiation and Matter
142510
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B As the electron and photon are having the same wavelength $\lambda$, the momentum of both of them will be the same. According to de-Broglie wavelength of a particle or photon, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
142500
The de-Broglie wavelength of $1 \mathrm{~kg}$ mass moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$, will be :
1 $6.626 \times 10^{-35} \mathrm{~m}$
2 $6.626 \times 10^{-34} \mathrm{~m}$
3 $6.626 \times 10^{-33} \mathrm{~m}$
4 none of these
Explanation:
A Given that, $\mathrm{m}=1 \mathrm{~kg}, \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\lambda=\frac{6.626 \times 10^{-34}}{1 \times 10}$ $\lambda=6.63 \times 10^{-35} \mathrm{~m}$
JCECE-2004
Dual nature of radiation and Matter
142502
If the momentum of a particle is doubled, then its de-Broglie wavelength will :
1 become four times
2 become half
3 become two times
4 remain unchanged
Explanation:
B From de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\mathrm{p}=\mathrm{mv})$ Hence if the momentum of a particle is doubled, value of $\lambda$ will be halved.
JCECE-2003
Dual nature of radiation and Matter
142509
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of $1 \mathrm{kV}$, its kinetic energy will be
1 $1840 \mathrm{keV}$
2 $1 / 1840 \mathrm{keV}$
3 $1 \mathrm{keV}$
4 $920 \mathrm{~V}$
Explanation:
C Given that, Mass of proton $=1840 \times$ mass of an electron Final kinetic energy - Initial kinetic $=$ Total change in energy. K.E $-0=$ Charge $\times$ potential difference $\mathrm{K} . \mathrm{E}=\mathrm{e} \times 1 \mathrm{kV}=1 \mathrm{keV}$
AIIMS-2003
Dual nature of radiation and Matter
142510
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B As the electron and photon are having the same wavelength $\lambda$, the momentum of both of them will be the same. According to de-Broglie wavelength of a particle or photon, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
142500
The de-Broglie wavelength of $1 \mathrm{~kg}$ mass moving with a velocity of $10 \mathrm{~m} / \mathrm{s}$, will be :
1 $6.626 \times 10^{-35} \mathrm{~m}$
2 $6.626 \times 10^{-34} \mathrm{~m}$
3 $6.626 \times 10^{-33} \mathrm{~m}$
4 none of these
Explanation:
A Given that, $\mathrm{m}=1 \mathrm{~kg}, \mathrm{v}=10 \mathrm{~m} / \mathrm{s}$ We know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}}$ $\lambda=\frac{6.626 \times 10^{-34}}{1 \times 10}$ $\lambda=6.63 \times 10^{-35} \mathrm{~m}$
JCECE-2004
Dual nature of radiation and Matter
142502
If the momentum of a particle is doubled, then its de-Broglie wavelength will :
1 become four times
2 become half
3 become two times
4 remain unchanged
Explanation:
B From de-Broglie wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \quad(\mathrm{p}=\mathrm{mv})$ Hence if the momentum of a particle is doubled, value of $\lambda$ will be halved.
JCECE-2003
Dual nature of radiation and Matter
142509
A proton is about 1840 times heavier than an electron. When it is accelerated by a potential difference of $1 \mathrm{kV}$, its kinetic energy will be
1 $1840 \mathrm{keV}$
2 $1 / 1840 \mathrm{keV}$
3 $1 \mathrm{keV}$
4 $920 \mathrm{~V}$
Explanation:
C Given that, Mass of proton $=1840 \times$ mass of an electron Final kinetic energy - Initial kinetic $=$ Total change in energy. K.E $-0=$ Charge $\times$ potential difference $\mathrm{K} . \mathrm{E}=\mathrm{e} \times 1 \mathrm{kV}=1 \mathrm{keV}$
AIIMS-2003
Dual nature of radiation and Matter
142510
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B As the electron and photon are having the same wavelength $\lambda$, the momentum of both of them will be the same. According to de-Broglie wavelength of a particle or photon, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$
