142560
Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $\left(\lambda_{0}\right)$ of the emitted $X$-rays is
A According to de-Brogile equation $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { and } \quad \sqrt{2 \mathrm{mE}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\mathrm{E} =\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}}}$ $\lambda =\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{~h}}$
JIPMER-2015
Dual nature of radiation and Matter
142561
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
1 25
2 75
3 60
4 50
Explanation:
B According to de-Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} 16(\mathrm{KE})}}=\frac{\mathrm{h}}{4 \sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\lambda_{1}}{4}$ Change in wavelength $=\lambda_{1}-\lambda_{2}$ $=\frac{3 \lambda_{1}}{4}$ $\% \text { Change }=\frac{\frac{3 \lambda_{1}}{4}}{\lambda_{1}} \times 100$ $=75 \%$
AIPMT - 2014
Dual nature of radiation and Matter
142563
The energy of a photon of light is $3 \mathrm{eV}$. Then the wavelength of photon must be
142566
A $200 \mathrm{~W}$ sodium street lamp emits yellow light of wavelength $0.6 \mu \mathrm{m}$. Assuming it to be $25 \%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
1 $1.5 \times 10^{20}$
2 $6 \times 10^{18}$
3 $62 \times 10^{20}$
4 $3 \times 10^{19}$
Explanation:
A Given, Power $(\mathrm{P})=200 \mathrm{~W}, \lambda=0.6 \mu \mathrm{m}=0.6 \times 10^{-6} \mathrm{~m}$ Effective power $(E)=200 \times \frac{25}{100}$ $=50 \mathrm{~W}$ Number of photon per sec. $\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P} \lambda}{\mathrm{hc}}$ $=\frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}}=1.5 \times 10^{20}$ photons per secss
142560
Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $\left(\lambda_{0}\right)$ of the emitted $X$-rays is
A According to de-Brogile equation $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { and } \quad \sqrt{2 \mathrm{mE}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\mathrm{E} =\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}}}$ $\lambda =\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{~h}}$
JIPMER-2015
Dual nature of radiation and Matter
142561
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
1 25
2 75
3 60
4 50
Explanation:
B According to de-Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} 16(\mathrm{KE})}}=\frac{\mathrm{h}}{4 \sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\lambda_{1}}{4}$ Change in wavelength $=\lambda_{1}-\lambda_{2}$ $=\frac{3 \lambda_{1}}{4}$ $\% \text { Change }=\frac{\frac{3 \lambda_{1}}{4}}{\lambda_{1}} \times 100$ $=75 \%$
AIPMT - 2014
Dual nature of radiation and Matter
142563
The energy of a photon of light is $3 \mathrm{eV}$. Then the wavelength of photon must be
142566
A $200 \mathrm{~W}$ sodium street lamp emits yellow light of wavelength $0.6 \mu \mathrm{m}$. Assuming it to be $25 \%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
1 $1.5 \times 10^{20}$
2 $6 \times 10^{18}$
3 $62 \times 10^{20}$
4 $3 \times 10^{19}$
Explanation:
A Given, Power $(\mathrm{P})=200 \mathrm{~W}, \lambda=0.6 \mu \mathrm{m}=0.6 \times 10^{-6} \mathrm{~m}$ Effective power $(E)=200 \times \frac{25}{100}$ $=50 \mathrm{~W}$ Number of photon per sec. $\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P} \lambda}{\mathrm{hc}}$ $=\frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}}=1.5 \times 10^{20}$ photons per secss
142560
Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $\left(\lambda_{0}\right)$ of the emitted $X$-rays is
A According to de-Brogile equation $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { and } \quad \sqrt{2 \mathrm{mE}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\mathrm{E} =\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}}}$ $\lambda =\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{~h}}$
JIPMER-2015
Dual nature of radiation and Matter
142561
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
1 25
2 75
3 60
4 50
Explanation:
B According to de-Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} 16(\mathrm{KE})}}=\frac{\mathrm{h}}{4 \sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\lambda_{1}}{4}$ Change in wavelength $=\lambda_{1}-\lambda_{2}$ $=\frac{3 \lambda_{1}}{4}$ $\% \text { Change }=\frac{\frac{3 \lambda_{1}}{4}}{\lambda_{1}} \times 100$ $=75 \%$
AIPMT - 2014
Dual nature of radiation and Matter
142563
The energy of a photon of light is $3 \mathrm{eV}$. Then the wavelength of photon must be
142566
A $200 \mathrm{~W}$ sodium street lamp emits yellow light of wavelength $0.6 \mu \mathrm{m}$. Assuming it to be $25 \%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
1 $1.5 \times 10^{20}$
2 $6 \times 10^{18}$
3 $62 \times 10^{20}$
4 $3 \times 10^{19}$
Explanation:
A Given, Power $(\mathrm{P})=200 \mathrm{~W}, \lambda=0.6 \mu \mathrm{m}=0.6 \times 10^{-6} \mathrm{~m}$ Effective power $(E)=200 \times \frac{25}{100}$ $=50 \mathrm{~W}$ Number of photon per sec. $\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P} \lambda}{\mathrm{hc}}$ $=\frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}}=1.5 \times 10^{20}$ photons per secss
142560
Electrons of mass $m$ with de-Broglie wavelength $\lambda$ fall on the target in an X-ray tube. The cutoff wavelength $\left(\lambda_{0}\right)$ of the emitted $X$-rays is
A According to de-Brogile equation $\lambda =\frac{\mathrm{h}}{\mathrm{p}}$ $\text { and } \quad \sqrt{2 \mathrm{mE}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{mE}}}$ $\mathrm{E} =\frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}$ $\lambda =\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} \times \frac{\mathrm{hc}}{\lambda_{\mathrm{o}}}}}$ $\lambda =\frac{2 \mathrm{mc} \lambda^{2}}{\mathrm{~h}}$
JIPMER-2015
Dual nature of radiation and Matter
142561
If the kinetic energy of the particle is increased to 16 times its previous value, the percentage change in the de-Broglie wavelength of the particle is
1 25
2 75
3 60
4 50
Explanation:
B According to de-Broglie wavelength $\lambda_{1}=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m}(\mathrm{KE})}}$ $\lambda_{2}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{~m} 16(\mathrm{KE})}}=\frac{\mathrm{h}}{4 \sqrt{2 \mathrm{~m}(\mathrm{KE})}}=\frac{\lambda_{1}}{4}$ Change in wavelength $=\lambda_{1}-\lambda_{2}$ $=\frac{3 \lambda_{1}}{4}$ $\% \text { Change }=\frac{\frac{3 \lambda_{1}}{4}}{\lambda_{1}} \times 100$ $=75 \%$
AIPMT - 2014
Dual nature of radiation and Matter
142563
The energy of a photon of light is $3 \mathrm{eV}$. Then the wavelength of photon must be
142566
A $200 \mathrm{~W}$ sodium street lamp emits yellow light of wavelength $0.6 \mu \mathrm{m}$. Assuming it to be $25 \%$ efficient in converting electrical energy to light, the number of photons of yellow light it emits per second is
1 $1.5 \times 10^{20}$
2 $6 \times 10^{18}$
3 $62 \times 10^{20}$
4 $3 \times 10^{19}$
Explanation:
A Given, Power $(\mathrm{P})=200 \mathrm{~W}, \lambda=0.6 \mu \mathrm{m}=0.6 \times 10^{-6} \mathrm{~m}$ Effective power $(E)=200 \times \frac{25}{100}$ $=50 \mathrm{~W}$ Number of photon per sec. $\frac{\mathrm{n}}{\mathrm{t}}=\frac{\mathrm{P} \lambda}{\mathrm{hc}}$ $=\frac{50 \times 0.6 \times 10^{-6}}{6.6 \times 10^{-34} \times 3 \times 10^{8}}$ $\frac{\mathrm{n}}{\mathrm{t}}=1.5 \times 10^{20}$ photons per secss
