142303
For a particle of mass $m$ moving with kinetic energy $E$, the de Broglie wavelength is
1 $\mathrm{h} / 2 \mathrm{mE}$
2 $\mathrm{h} \sqrt{2 \mathrm{mE}}$
3 $\mathrm{h} / \sqrt{2 \mathrm{mE}}$
4 $\mathrm{h} \sqrt{2} / \mathrm{mE}$
Explanation:
C Given, mass $=\mathrm{m}$ Kinetic energy $=\mathrm{E}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ And $\quad E=\frac{1}{2} m v^{2}$ Momentum $(\mathrm{p})=\mathrm{mv}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ Then, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where, $\mathrm{h}$ is the Planck's constant and $\mathrm{E}$ is kinetic energy.
SRMJEEE - 2015
Dual nature of radiation and Matter
142304
If mean wavelength of light radiated by $100 \mathrm{~W}$ lamp is $5000 \AA$, the number of photons radiated per second would be
1 $2.5 \times 10^{20}$
2 $2.5 \times 10^{22}$
3 $3 \times 10^{23}$
4 $5 \times 10^{17}$
Explanation:
A Given that, Power $(\mathrm{P})=100 \mathrm{~W}$ Wavelength of light $(\lambda)=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ Power $(P)=\frac{W}{t}=\frac{n E}{t}=\frac{n}{t} E$ $P=\frac{n}{t} E$ $P=\frac{n}{t} \frac{h c}{\lambda}$ $\frac{n}{t}=\frac{P \times \lambda}{h c}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{100 \times 5000 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=2.5 \times 10^{20}$
SRMJEEE - 2009
Dual nature of radiation and Matter
142305
Wavelength of a $1 \mathrm{keV}$ photon is $1.24 \times 10^{-9} \mathrm{~m}$. What is the frequency of $1 \mathrm{MeV}$ photon?
142303
For a particle of mass $m$ moving with kinetic energy $E$, the de Broglie wavelength is
1 $\mathrm{h} / 2 \mathrm{mE}$
2 $\mathrm{h} \sqrt{2 \mathrm{mE}}$
3 $\mathrm{h} / \sqrt{2 \mathrm{mE}}$
4 $\mathrm{h} \sqrt{2} / \mathrm{mE}$
Explanation:
C Given, mass $=\mathrm{m}$ Kinetic energy $=\mathrm{E}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ And $\quad E=\frac{1}{2} m v^{2}$ Momentum $(\mathrm{p})=\mathrm{mv}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ Then, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where, $\mathrm{h}$ is the Planck's constant and $\mathrm{E}$ is kinetic energy.
SRMJEEE - 2015
Dual nature of radiation and Matter
142304
If mean wavelength of light radiated by $100 \mathrm{~W}$ lamp is $5000 \AA$, the number of photons radiated per second would be
1 $2.5 \times 10^{20}$
2 $2.5 \times 10^{22}$
3 $3 \times 10^{23}$
4 $5 \times 10^{17}$
Explanation:
A Given that, Power $(\mathrm{P})=100 \mathrm{~W}$ Wavelength of light $(\lambda)=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ Power $(P)=\frac{W}{t}=\frac{n E}{t}=\frac{n}{t} E$ $P=\frac{n}{t} E$ $P=\frac{n}{t} \frac{h c}{\lambda}$ $\frac{n}{t}=\frac{P \times \lambda}{h c}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{100 \times 5000 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=2.5 \times 10^{20}$
SRMJEEE - 2009
Dual nature of radiation and Matter
142305
Wavelength of a $1 \mathrm{keV}$ photon is $1.24 \times 10^{-9} \mathrm{~m}$. What is the frequency of $1 \mathrm{MeV}$ photon?
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142303
For a particle of mass $m$ moving with kinetic energy $E$, the de Broglie wavelength is
1 $\mathrm{h} / 2 \mathrm{mE}$
2 $\mathrm{h} \sqrt{2 \mathrm{mE}}$
3 $\mathrm{h} / \sqrt{2 \mathrm{mE}}$
4 $\mathrm{h} \sqrt{2} / \mathrm{mE}$
Explanation:
C Given, mass $=\mathrm{m}$ Kinetic energy $=\mathrm{E}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ And $\quad E=\frac{1}{2} m v^{2}$ Momentum $(\mathrm{p})=\mathrm{mv}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ Then, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where, $\mathrm{h}$ is the Planck's constant and $\mathrm{E}$ is kinetic energy.
SRMJEEE - 2015
Dual nature of radiation and Matter
142304
If mean wavelength of light radiated by $100 \mathrm{~W}$ lamp is $5000 \AA$, the number of photons radiated per second would be
1 $2.5 \times 10^{20}$
2 $2.5 \times 10^{22}$
3 $3 \times 10^{23}$
4 $5 \times 10^{17}$
Explanation:
A Given that, Power $(\mathrm{P})=100 \mathrm{~W}$ Wavelength of light $(\lambda)=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ Power $(P)=\frac{W}{t}=\frac{n E}{t}=\frac{n}{t} E$ $P=\frac{n}{t} E$ $P=\frac{n}{t} \frac{h c}{\lambda}$ $\frac{n}{t}=\frac{P \times \lambda}{h c}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{100 \times 5000 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=2.5 \times 10^{20}$
SRMJEEE - 2009
Dual nature of radiation and Matter
142305
Wavelength of a $1 \mathrm{keV}$ photon is $1.24 \times 10^{-9} \mathrm{~m}$. What is the frequency of $1 \mathrm{MeV}$ photon?
142303
For a particle of mass $m$ moving with kinetic energy $E$, the de Broglie wavelength is
1 $\mathrm{h} / 2 \mathrm{mE}$
2 $\mathrm{h} \sqrt{2 \mathrm{mE}}$
3 $\mathrm{h} / \sqrt{2 \mathrm{mE}}$
4 $\mathrm{h} \sqrt{2} / \mathrm{mE}$
Explanation:
C Given, mass $=\mathrm{m}$ Kinetic energy $=\mathrm{E}$ We know that, de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}$ And $\quad E=\frac{1}{2} m v^{2}$ Momentum $(\mathrm{p})=\mathrm{mv}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}}$ Then, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ Where, $\mathrm{h}$ is the Planck's constant and $\mathrm{E}$ is kinetic energy.
SRMJEEE - 2015
Dual nature of radiation and Matter
142304
If mean wavelength of light radiated by $100 \mathrm{~W}$ lamp is $5000 \AA$, the number of photons radiated per second would be
1 $2.5 \times 10^{20}$
2 $2.5 \times 10^{22}$
3 $3 \times 10^{23}$
4 $5 \times 10^{17}$
Explanation:
A Given that, Power $(\mathrm{P})=100 \mathrm{~W}$ Wavelength of light $(\lambda)=5000 \AA=5000 \times 10^{-10} \mathrm{~m}$ Power $(P)=\frac{W}{t}=\frac{n E}{t}=\frac{n}{t} E$ $P=\frac{n}{t} E$ $P=\frac{n}{t} \frac{h c}{\lambda}$ $\frac{n}{t}=\frac{P \times \lambda}{h c}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=\frac{100 \times 5000 \times 10^{-10}}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\left(\frac{\mathrm{n}}{\mathrm{t}}\right)=2.5 \times 10^{20}$
SRMJEEE - 2009
Dual nature of radiation and Matter
142305
Wavelength of a $1 \mathrm{keV}$ photon is $1.24 \times 10^{-9} \mathrm{~m}$. What is the frequency of $1 \mathrm{MeV}$ photon?