NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142441
A particle of mass $6 \mathrm{~m}$ at rest decays into two particles of masses $2 \mathrm{~m}$ and $4 \mathrm{~m}$ having nonzero velocities. What will be the ratio of the de Broglie wavelengths of two particles?
1 2
2 $1 / 2$
3 1
4 $1 / 4$
Explanation:
C By conservation of linear momentum $\mathrm{Mv}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ Since, $\mathrm{M}$ is initially in rest $\mathrm{so} \mathrm{v}=0$ $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$ $\mathrm{~m}_{1} \mathrm{v}_{1}=-\mathrm{m}_{2} \mathrm{v}_{2}$ That's means the magnitude of momentum of both mass or equal so de- Broglie wavelength, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}} .$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}}{\mathrm{p}}=1$
J and K CET-2019
Dual nature of radiation and Matter
142442
The wavelength of a light coming from a sodium source is $645 \mathrm{~nm}$ in air. What will be its wavelength in water, if the refractive index of water is 1.33 ?
1 $645 \mathrm{~nm}$
2 $485 \mathrm{~nm}$
3 $408 \mathrm{~nm}$
4 $480 \mathrm{~nm}$
Explanation:
B Given that, $\lambda_{\mathrm{a}}=645 \mathrm{~nm}$ $\mu_{\mathrm{a}}=1$ $\mu_{\mathrm{w}}=1.33$ $\lambda_{\mathrm{w}}=?$ We know that, $\lambda \propto \frac{1}{\mu}$ Then, $\quad \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}=\frac{\lambda_{\mathrm{w}}}{\lambda_{\mathrm{a}}}$ $\lambda_{\mathrm{w}}=\lambda_{\mathrm{a}} \times \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}$ $\lambda_{\mathrm{w}}=645 \times \frac{1}{1.33}=484.96$ So, $\quad \lambda_{\mathrm{w}} \square 485 \mathrm{~nm}$
J and K CET-2019
Dual nature of radiation and Matter
142445
To increase de-Broglie wavelength of an electron from $0.5 \times 10^{-10} \mathrm{~m}$ to $10^{-10} \mathrm{~m}$, its energy should be
1 increased to 4 times
2 halved
3 doubled
4 decreased to fourth part
Explanation:
D As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\mathrm{K}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ According to the question, If $\lambda$ becomes $2 \lambda$ Then, $K=\frac{h^{2}}{2 m\left(4 \lambda^{2}\right)}$ $\mathrm{K}=\frac{1}{4}\left[\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}\right]$ So, its energy should be decreased to fourth part.
GUJCET 2019
Dual nature of radiation and Matter
142446
An $\alpha$ - particle moves in a circular path of radius $1 \mathrm{~cm}$ in a uniform magnetic field of $0.125 \mathrm{~T}$. The de-Broglie wavelength associated with the $\alpha$-particle is
1 $1.65 \times 10^{-12} \mathrm{~m}$
2 $3.3 \times 10^{-12} \mathrm{~m}$
3 $4.95 \times 10^{-12} \mathrm{~m}$
4 $6.6 \times 10^{-12} \mathrm{~m}$
Explanation:
A Given that, $\mathrm{q}_{\alpha}=2 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{r}=1 \times 10^{-2} \mathrm{~m}$ $\mathrm{B}=0.125 \mathrm{~T}$ Magnetic force $\left(\mathrm{F}_{\mathrm{m}}\right)=\mathrm{B} \cdot \mathrm{q}_{\alpha} \cdot \mathrm{V}$ Centripetal force $\left(F_{c}\right)=\frac{m v^{2}}{r}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bq}_{\alpha} \cdot \mathrm{V}$ $\mathrm{mv}=\mathrm{Bq}_{\alpha} \mathrm{r}$ According to the de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $\mathrm{mv}$ from equation (iii) in equation (iv), we get - $\lambda=\frac{\mathrm{h}}{\mathrm{Bq}_{\alpha} \mathrm{r}}$ $\lambda=\frac{6.6 \times 10^{-34}}{0.125 \times 3.2 \times 10^{-19} \times 1 \times 10^{-2}}$ $\lambda=1.65 \times 10^{-12} \mathrm{~m}$
142441
A particle of mass $6 \mathrm{~m}$ at rest decays into two particles of masses $2 \mathrm{~m}$ and $4 \mathrm{~m}$ having nonzero velocities. What will be the ratio of the de Broglie wavelengths of two particles?
1 2
2 $1 / 2$
3 1
4 $1 / 4$
Explanation:
C By conservation of linear momentum $\mathrm{Mv}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ Since, $\mathrm{M}$ is initially in rest $\mathrm{so} \mathrm{v}=0$ $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$ $\mathrm{~m}_{1} \mathrm{v}_{1}=-\mathrm{m}_{2} \mathrm{v}_{2}$ That's means the magnitude of momentum of both mass or equal so de- Broglie wavelength, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}} .$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}}{\mathrm{p}}=1$
J and K CET-2019
Dual nature of radiation and Matter
142442
The wavelength of a light coming from a sodium source is $645 \mathrm{~nm}$ in air. What will be its wavelength in water, if the refractive index of water is 1.33 ?
1 $645 \mathrm{~nm}$
2 $485 \mathrm{~nm}$
3 $408 \mathrm{~nm}$
4 $480 \mathrm{~nm}$
Explanation:
B Given that, $\lambda_{\mathrm{a}}=645 \mathrm{~nm}$ $\mu_{\mathrm{a}}=1$ $\mu_{\mathrm{w}}=1.33$ $\lambda_{\mathrm{w}}=?$ We know that, $\lambda \propto \frac{1}{\mu}$ Then, $\quad \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}=\frac{\lambda_{\mathrm{w}}}{\lambda_{\mathrm{a}}}$ $\lambda_{\mathrm{w}}=\lambda_{\mathrm{a}} \times \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}$ $\lambda_{\mathrm{w}}=645 \times \frac{1}{1.33}=484.96$ So, $\quad \lambda_{\mathrm{w}} \square 485 \mathrm{~nm}$
J and K CET-2019
Dual nature of radiation and Matter
142445
To increase de-Broglie wavelength of an electron from $0.5 \times 10^{-10} \mathrm{~m}$ to $10^{-10} \mathrm{~m}$, its energy should be
1 increased to 4 times
2 halved
3 doubled
4 decreased to fourth part
Explanation:
D As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\mathrm{K}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ According to the question, If $\lambda$ becomes $2 \lambda$ Then, $K=\frac{h^{2}}{2 m\left(4 \lambda^{2}\right)}$ $\mathrm{K}=\frac{1}{4}\left[\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}\right]$ So, its energy should be decreased to fourth part.
GUJCET 2019
Dual nature of radiation and Matter
142446
An $\alpha$ - particle moves in a circular path of radius $1 \mathrm{~cm}$ in a uniform magnetic field of $0.125 \mathrm{~T}$. The de-Broglie wavelength associated with the $\alpha$-particle is
1 $1.65 \times 10^{-12} \mathrm{~m}$
2 $3.3 \times 10^{-12} \mathrm{~m}$
3 $4.95 \times 10^{-12} \mathrm{~m}$
4 $6.6 \times 10^{-12} \mathrm{~m}$
Explanation:
A Given that, $\mathrm{q}_{\alpha}=2 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{r}=1 \times 10^{-2} \mathrm{~m}$ $\mathrm{B}=0.125 \mathrm{~T}$ Magnetic force $\left(\mathrm{F}_{\mathrm{m}}\right)=\mathrm{B} \cdot \mathrm{q}_{\alpha} \cdot \mathrm{V}$ Centripetal force $\left(F_{c}\right)=\frac{m v^{2}}{r}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bq}_{\alpha} \cdot \mathrm{V}$ $\mathrm{mv}=\mathrm{Bq}_{\alpha} \mathrm{r}$ According to the de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $\mathrm{mv}$ from equation (iii) in equation (iv), we get - $\lambda=\frac{\mathrm{h}}{\mathrm{Bq}_{\alpha} \mathrm{r}}$ $\lambda=\frac{6.6 \times 10^{-34}}{0.125 \times 3.2 \times 10^{-19} \times 1 \times 10^{-2}}$ $\lambda=1.65 \times 10^{-12} \mathrm{~m}$
142441
A particle of mass $6 \mathrm{~m}$ at rest decays into two particles of masses $2 \mathrm{~m}$ and $4 \mathrm{~m}$ having nonzero velocities. What will be the ratio of the de Broglie wavelengths of two particles?
1 2
2 $1 / 2$
3 1
4 $1 / 4$
Explanation:
C By conservation of linear momentum $\mathrm{Mv}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ Since, $\mathrm{M}$ is initially in rest $\mathrm{so} \mathrm{v}=0$ $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$ $\mathrm{~m}_{1} \mathrm{v}_{1}=-\mathrm{m}_{2} \mathrm{v}_{2}$ That's means the magnitude of momentum of both mass or equal so de- Broglie wavelength, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}} .$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}}{\mathrm{p}}=1$
J and K CET-2019
Dual nature of radiation and Matter
142442
The wavelength of a light coming from a sodium source is $645 \mathrm{~nm}$ in air. What will be its wavelength in water, if the refractive index of water is 1.33 ?
1 $645 \mathrm{~nm}$
2 $485 \mathrm{~nm}$
3 $408 \mathrm{~nm}$
4 $480 \mathrm{~nm}$
Explanation:
B Given that, $\lambda_{\mathrm{a}}=645 \mathrm{~nm}$ $\mu_{\mathrm{a}}=1$ $\mu_{\mathrm{w}}=1.33$ $\lambda_{\mathrm{w}}=?$ We know that, $\lambda \propto \frac{1}{\mu}$ Then, $\quad \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}=\frac{\lambda_{\mathrm{w}}}{\lambda_{\mathrm{a}}}$ $\lambda_{\mathrm{w}}=\lambda_{\mathrm{a}} \times \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}$ $\lambda_{\mathrm{w}}=645 \times \frac{1}{1.33}=484.96$ So, $\quad \lambda_{\mathrm{w}} \square 485 \mathrm{~nm}$
J and K CET-2019
Dual nature of radiation and Matter
142445
To increase de-Broglie wavelength of an electron from $0.5 \times 10^{-10} \mathrm{~m}$ to $10^{-10} \mathrm{~m}$, its energy should be
1 increased to 4 times
2 halved
3 doubled
4 decreased to fourth part
Explanation:
D As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\mathrm{K}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ According to the question, If $\lambda$ becomes $2 \lambda$ Then, $K=\frac{h^{2}}{2 m\left(4 \lambda^{2}\right)}$ $\mathrm{K}=\frac{1}{4}\left[\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}\right]$ So, its energy should be decreased to fourth part.
GUJCET 2019
Dual nature of radiation and Matter
142446
An $\alpha$ - particle moves in a circular path of radius $1 \mathrm{~cm}$ in a uniform magnetic field of $0.125 \mathrm{~T}$. The de-Broglie wavelength associated with the $\alpha$-particle is
1 $1.65 \times 10^{-12} \mathrm{~m}$
2 $3.3 \times 10^{-12} \mathrm{~m}$
3 $4.95 \times 10^{-12} \mathrm{~m}$
4 $6.6 \times 10^{-12} \mathrm{~m}$
Explanation:
A Given that, $\mathrm{q}_{\alpha}=2 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{r}=1 \times 10^{-2} \mathrm{~m}$ $\mathrm{B}=0.125 \mathrm{~T}$ Magnetic force $\left(\mathrm{F}_{\mathrm{m}}\right)=\mathrm{B} \cdot \mathrm{q}_{\alpha} \cdot \mathrm{V}$ Centripetal force $\left(F_{c}\right)=\frac{m v^{2}}{r}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bq}_{\alpha} \cdot \mathrm{V}$ $\mathrm{mv}=\mathrm{Bq}_{\alpha} \mathrm{r}$ According to the de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $\mathrm{mv}$ from equation (iii) in equation (iv), we get - $\lambda=\frac{\mathrm{h}}{\mathrm{Bq}_{\alpha} \mathrm{r}}$ $\lambda=\frac{6.6 \times 10^{-34}}{0.125 \times 3.2 \times 10^{-19} \times 1 \times 10^{-2}}$ $\lambda=1.65 \times 10^{-12} \mathrm{~m}$
142441
A particle of mass $6 \mathrm{~m}$ at rest decays into two particles of masses $2 \mathrm{~m}$ and $4 \mathrm{~m}$ having nonzero velocities. What will be the ratio of the de Broglie wavelengths of two particles?
1 2
2 $1 / 2$
3 1
4 $1 / 4$
Explanation:
C By conservation of linear momentum $\mathrm{Mv}=\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}$ Since, $\mathrm{M}$ is initially in rest $\mathrm{so} \mathrm{v}=0$ $\mathrm{m}_{1} \mathrm{v}_{1}+\mathrm{m}_{2} \mathrm{v}_{2}=0$ $\mathrm{~m}_{1} \mathrm{v}_{1}=-\mathrm{m}_{2} \mathrm{v}_{2}$ That's means the magnitude of momentum of both mass or equal so de- Broglie wavelength, $\mathrm{p}_{1}=\mathrm{p}_{2}=\mathrm{p}$ $\lambda \propto \frac{1}{\mathrm{p}}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}} .$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{p}}{\mathrm{p}}=1$
J and K CET-2019
Dual nature of radiation and Matter
142442
The wavelength of a light coming from a sodium source is $645 \mathrm{~nm}$ in air. What will be its wavelength in water, if the refractive index of water is 1.33 ?
1 $645 \mathrm{~nm}$
2 $485 \mathrm{~nm}$
3 $408 \mathrm{~nm}$
4 $480 \mathrm{~nm}$
Explanation:
B Given that, $\lambda_{\mathrm{a}}=645 \mathrm{~nm}$ $\mu_{\mathrm{a}}=1$ $\mu_{\mathrm{w}}=1.33$ $\lambda_{\mathrm{w}}=?$ We know that, $\lambda \propto \frac{1}{\mu}$ Then, $\quad \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}=\frac{\lambda_{\mathrm{w}}}{\lambda_{\mathrm{a}}}$ $\lambda_{\mathrm{w}}=\lambda_{\mathrm{a}} \times \frac{\mu_{\mathrm{a}}}{\mu_{\mathrm{w}}}$ $\lambda_{\mathrm{w}}=645 \times \frac{1}{1.33}=484.96$ So, $\quad \lambda_{\mathrm{w}} \square 485 \mathrm{~nm}$
J and K CET-2019
Dual nature of radiation and Matter
142445
To increase de-Broglie wavelength of an electron from $0.5 \times 10^{-10} \mathrm{~m}$ to $10^{-10} \mathrm{~m}$, its energy should be
1 increased to 4 times
2 halved
3 doubled
4 decreased to fourth part
Explanation:
D As we know that, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\mathrm{K}=\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}$ According to the question, If $\lambda$ becomes $2 \lambda$ Then, $K=\frac{h^{2}}{2 m\left(4 \lambda^{2}\right)}$ $\mathrm{K}=\frac{1}{4}\left[\frac{\mathrm{h}^{2}}{2 \mathrm{~m} \lambda^{2}}\right]$ So, its energy should be decreased to fourth part.
GUJCET 2019
Dual nature of radiation and Matter
142446
An $\alpha$ - particle moves in a circular path of radius $1 \mathrm{~cm}$ in a uniform magnetic field of $0.125 \mathrm{~T}$. The de-Broglie wavelength associated with the $\alpha$-particle is
1 $1.65 \times 10^{-12} \mathrm{~m}$
2 $3.3 \times 10^{-12} \mathrm{~m}$
3 $4.95 \times 10^{-12} \mathrm{~m}$
4 $6.6 \times 10^{-12} \mathrm{~m}$
Explanation:
A Given that, $\mathrm{q}_{\alpha}=2 \times 1.6 \times 10^{-19} \mathrm{C}, \mathrm{r}=1 \times 10^{-2} \mathrm{~m}$ $\mathrm{B}=0.125 \mathrm{~T}$ Magnetic force $\left(\mathrm{F}_{\mathrm{m}}\right)=\mathrm{B} \cdot \mathrm{q}_{\alpha} \cdot \mathrm{V}$ Centripetal force $\left(F_{c}\right)=\frac{m v^{2}}{r}$ From equation (i) and equation (ii), we get - $\frac{\mathrm{mv}^{2}}{\mathrm{r}}=\mathrm{Bq}_{\alpha} \cdot \mathrm{V}$ $\mathrm{mv}=\mathrm{Bq}_{\alpha} \mathrm{r}$ According to the de-Broglie wavelength- $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\mathrm{mv}}$ Putting the value of $\mathrm{mv}$ from equation (iii) in equation (iv), we get - $\lambda=\frac{\mathrm{h}}{\mathrm{Bq}_{\alpha} \mathrm{r}}$ $\lambda=\frac{6.6 \times 10^{-34}}{0.125 \times 3.2 \times 10^{-19} \times 1 \times 10^{-2}}$ $\lambda=1.65 \times 10^{-12} \mathrm{~m}$
