B Given that, $\mathrm{n}_{\mathrm{R}}=$ red bulb, $\mathrm{n}_{\mathrm{V}}=$ violet bulb We know that, energy possessed by a photon - $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ If the power of each photon is ' $\mathrm{P}$ ' then energy given out in $t$ second is equal to 'Pt'. Let the number of photons be 'n'. $\text { Then, } \mathrm{n}=\frac{\mathrm{Pt}}{\mathrm{E}}$ $\mathrm{n} =\frac{\mathrm{Pt}}{(\mathrm{hc} / \lambda)}$ $\mathrm{n} =\frac{\mathrm{Pt} \lambda}{\mathrm{hc}}$ For red light, $\mathrm{n}_{\mathrm{R}}=\frac{\mathrm{Pt} \lambda_{\mathrm{R}}}{\mathrm{hc}}$ For violet light, $\mathrm{n}_{\mathrm{v}}=\frac{\mathrm{Pt} \lambda_{\mathrm{v}}}{\mathrm{hc}}$ From equation (i) and (ii), we get - $\frac{\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{V}}}=\frac{\lambda_{\mathrm{R}}}{\lambda_{\mathrm{V}}}$ As we know that, $\lambda_{\mathrm{R}}>\lambda_{\mathrm{V}}$ Then $n_{R}$ is less than $n_{V}$. $\text { So, } \quad \mathrm{n}_{\mathrm{R}}>\mathrm{n}_{\mathrm{V}}$
MHT-CET 2006
Dual nature of radiation and Matter
142377
Which is the incorrect statement of the following?
1 Photon is a particle with zero rest mass
2 Photon is a particle with zero momentum
3 Photon travel with velocity of light in vacuum
4 Photon even feel the pull of gravity
Explanation:
B Photon is a minute energy packet electromagnetic radiation. - Each photon has momentum $(p)=\frac{h v}{c}$, energy $(E)=$ $h v$ and speed (c) the speed of light. - Photons have zero resting mass. - The total momentum and total energy are conserved in photon-particle collision. - All photon of the light of a particular frequency and wavelength have the same energy.
VITEEE-2006
Dual nature of radiation and Matter
142382
Number of ejected photoelectron increases with increase
1 in intensity of light
2 in wavelength of light
3 in frequency of light
4 Never
Explanation:
A As we know that, Intensity is the quantity of energy with the wave conveys per unit times across the surface of the per unit area. If frequency is constant, intensity increases that will increase in energy thus increase in number of ejected electron. Hence, Number of ejected photoelectron increases with increase in intensity of light.
AIPMT - 1993
Dual nature of radiation and Matter
142358
Which of the following statements is NOT true?
1 Electromagnetic radiation is made up of particles called photons
2 Each photon moves with the speed of light
3 Photon energy is dependent on the intensity of radiation
4 Photons are not deflected by electric and magnetic field
Explanation:
C Photon's energy depends on wavelength in such a way that the energy of the photon is inversely proportional to the wavelength. The higher the photon energy frequency, the higher its energy. It does not depend on the intensity of radiation.
B Given that, $\mathrm{n}_{\mathrm{R}}=$ red bulb, $\mathrm{n}_{\mathrm{V}}=$ violet bulb We know that, energy possessed by a photon - $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ If the power of each photon is ' $\mathrm{P}$ ' then energy given out in $t$ second is equal to 'Pt'. Let the number of photons be 'n'. $\text { Then, } \mathrm{n}=\frac{\mathrm{Pt}}{\mathrm{E}}$ $\mathrm{n} =\frac{\mathrm{Pt}}{(\mathrm{hc} / \lambda)}$ $\mathrm{n} =\frac{\mathrm{Pt} \lambda}{\mathrm{hc}}$ For red light, $\mathrm{n}_{\mathrm{R}}=\frac{\mathrm{Pt} \lambda_{\mathrm{R}}}{\mathrm{hc}}$ For violet light, $\mathrm{n}_{\mathrm{v}}=\frac{\mathrm{Pt} \lambda_{\mathrm{v}}}{\mathrm{hc}}$ From equation (i) and (ii), we get - $\frac{\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{V}}}=\frac{\lambda_{\mathrm{R}}}{\lambda_{\mathrm{V}}}$ As we know that, $\lambda_{\mathrm{R}}>\lambda_{\mathrm{V}}$ Then $n_{R}$ is less than $n_{V}$. $\text { So, } \quad \mathrm{n}_{\mathrm{R}}>\mathrm{n}_{\mathrm{V}}$
MHT-CET 2006
Dual nature of radiation and Matter
142377
Which is the incorrect statement of the following?
1 Photon is a particle with zero rest mass
2 Photon is a particle with zero momentum
3 Photon travel with velocity of light in vacuum
4 Photon even feel the pull of gravity
Explanation:
B Photon is a minute energy packet electromagnetic radiation. - Each photon has momentum $(p)=\frac{h v}{c}$, energy $(E)=$ $h v$ and speed (c) the speed of light. - Photons have zero resting mass. - The total momentum and total energy are conserved in photon-particle collision. - All photon of the light of a particular frequency and wavelength have the same energy.
VITEEE-2006
Dual nature of radiation and Matter
142382
Number of ejected photoelectron increases with increase
1 in intensity of light
2 in wavelength of light
3 in frequency of light
4 Never
Explanation:
A As we know that, Intensity is the quantity of energy with the wave conveys per unit times across the surface of the per unit area. If frequency is constant, intensity increases that will increase in energy thus increase in number of ejected electron. Hence, Number of ejected photoelectron increases with increase in intensity of light.
AIPMT - 1993
Dual nature of radiation and Matter
142358
Which of the following statements is NOT true?
1 Electromagnetic radiation is made up of particles called photons
2 Each photon moves with the speed of light
3 Photon energy is dependent on the intensity of radiation
4 Photons are not deflected by electric and magnetic field
Explanation:
C Photon's energy depends on wavelength in such a way that the energy of the photon is inversely proportional to the wavelength. The higher the photon energy frequency, the higher its energy. It does not depend on the intensity of radiation.
B Given that, $\mathrm{n}_{\mathrm{R}}=$ red bulb, $\mathrm{n}_{\mathrm{V}}=$ violet bulb We know that, energy possessed by a photon - $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ If the power of each photon is ' $\mathrm{P}$ ' then energy given out in $t$ second is equal to 'Pt'. Let the number of photons be 'n'. $\text { Then, } \mathrm{n}=\frac{\mathrm{Pt}}{\mathrm{E}}$ $\mathrm{n} =\frac{\mathrm{Pt}}{(\mathrm{hc} / \lambda)}$ $\mathrm{n} =\frac{\mathrm{Pt} \lambda}{\mathrm{hc}}$ For red light, $\mathrm{n}_{\mathrm{R}}=\frac{\mathrm{Pt} \lambda_{\mathrm{R}}}{\mathrm{hc}}$ For violet light, $\mathrm{n}_{\mathrm{v}}=\frac{\mathrm{Pt} \lambda_{\mathrm{v}}}{\mathrm{hc}}$ From equation (i) and (ii), we get - $\frac{\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{V}}}=\frac{\lambda_{\mathrm{R}}}{\lambda_{\mathrm{V}}}$ As we know that, $\lambda_{\mathrm{R}}>\lambda_{\mathrm{V}}$ Then $n_{R}$ is less than $n_{V}$. $\text { So, } \quad \mathrm{n}_{\mathrm{R}}>\mathrm{n}_{\mathrm{V}}$
MHT-CET 2006
Dual nature of radiation and Matter
142377
Which is the incorrect statement of the following?
1 Photon is a particle with zero rest mass
2 Photon is a particle with zero momentum
3 Photon travel with velocity of light in vacuum
4 Photon even feel the pull of gravity
Explanation:
B Photon is a minute energy packet electromagnetic radiation. - Each photon has momentum $(p)=\frac{h v}{c}$, energy $(E)=$ $h v$ and speed (c) the speed of light. - Photons have zero resting mass. - The total momentum and total energy are conserved in photon-particle collision. - All photon of the light of a particular frequency and wavelength have the same energy.
VITEEE-2006
Dual nature of radiation and Matter
142382
Number of ejected photoelectron increases with increase
1 in intensity of light
2 in wavelength of light
3 in frequency of light
4 Never
Explanation:
A As we know that, Intensity is the quantity of energy with the wave conveys per unit times across the surface of the per unit area. If frequency is constant, intensity increases that will increase in energy thus increase in number of ejected electron. Hence, Number of ejected photoelectron increases with increase in intensity of light.
AIPMT - 1993
Dual nature of radiation and Matter
142358
Which of the following statements is NOT true?
1 Electromagnetic radiation is made up of particles called photons
2 Each photon moves with the speed of light
3 Photon energy is dependent on the intensity of radiation
4 Photons are not deflected by electric and magnetic field
Explanation:
C Photon's energy depends on wavelength in such a way that the energy of the photon is inversely proportional to the wavelength. The higher the photon energy frequency, the higher its energy. It does not depend on the intensity of radiation.
B Given that, $\mathrm{n}_{\mathrm{R}}=$ red bulb, $\mathrm{n}_{\mathrm{V}}=$ violet bulb We know that, energy possessed by a photon - $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ If the power of each photon is ' $\mathrm{P}$ ' then energy given out in $t$ second is equal to 'Pt'. Let the number of photons be 'n'. $\text { Then, } \mathrm{n}=\frac{\mathrm{Pt}}{\mathrm{E}}$ $\mathrm{n} =\frac{\mathrm{Pt}}{(\mathrm{hc} / \lambda)}$ $\mathrm{n} =\frac{\mathrm{Pt} \lambda}{\mathrm{hc}}$ For red light, $\mathrm{n}_{\mathrm{R}}=\frac{\mathrm{Pt} \lambda_{\mathrm{R}}}{\mathrm{hc}}$ For violet light, $\mathrm{n}_{\mathrm{v}}=\frac{\mathrm{Pt} \lambda_{\mathrm{v}}}{\mathrm{hc}}$ From equation (i) and (ii), we get - $\frac{\mathrm{n}_{\mathrm{R}}}{\mathrm{n}_{\mathrm{V}}}=\frac{\lambda_{\mathrm{R}}}{\lambda_{\mathrm{V}}}$ As we know that, $\lambda_{\mathrm{R}}>\lambda_{\mathrm{V}}$ Then $n_{R}$ is less than $n_{V}$. $\text { So, } \quad \mathrm{n}_{\mathrm{R}}>\mathrm{n}_{\mathrm{V}}$
MHT-CET 2006
Dual nature of radiation and Matter
142377
Which is the incorrect statement of the following?
1 Photon is a particle with zero rest mass
2 Photon is a particle with zero momentum
3 Photon travel with velocity of light in vacuum
4 Photon even feel the pull of gravity
Explanation:
B Photon is a minute energy packet electromagnetic radiation. - Each photon has momentum $(p)=\frac{h v}{c}$, energy $(E)=$ $h v$ and speed (c) the speed of light. - Photons have zero resting mass. - The total momentum and total energy are conserved in photon-particle collision. - All photon of the light of a particular frequency and wavelength have the same energy.
VITEEE-2006
Dual nature of radiation and Matter
142382
Number of ejected photoelectron increases with increase
1 in intensity of light
2 in wavelength of light
3 in frequency of light
4 Never
Explanation:
A As we know that, Intensity is the quantity of energy with the wave conveys per unit times across the surface of the per unit area. If frequency is constant, intensity increases that will increase in energy thus increase in number of ejected electron. Hence, Number of ejected photoelectron increases with increase in intensity of light.
AIPMT - 1993
Dual nature of radiation and Matter
142358
Which of the following statements is NOT true?
1 Electromagnetic radiation is made up of particles called photons
2 Each photon moves with the speed of light
3 Photon energy is dependent on the intensity of radiation
4 Photons are not deflected by electric and magnetic field
Explanation:
C Photon's energy depends on wavelength in such a way that the energy of the photon is inversely proportional to the wavelength. The higher the photon energy frequency, the higher its energy. It does not depend on the intensity of radiation.
