142366
It the energy of photons corresponding to wavelength of $6000 \AA$ is $3.2 \times 10^{-19} \mathrm{~J}$. The photon energy for wavelength of $4000 \AA$ will be
1 $4.44 \times 10^{-19} \mathrm{~J}$
2 $2.22 \times 10^{-19} \mathrm{~J}$
3 $1.11 \times 10^{-19} \mathrm{~J}$
4 $4.80 \times 10^{-19} \mathrm{~J}$
Explanation:
D Given that, $\lambda_{1}=6000 \AA, \lambda_{2}=4000 \AA, \mathrm{E}_{1}=$ $3.2 \times 10^{-19} \mathrm{~J}$ We know that, $E=\frac{h c}{\lambda}$ $\text { So, } \quad E \propto \frac{1}{\lambda}$ Therefore, $\frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{4000}{6000}=\frac{2}{3}$ $\mathrm{E}_{2}=\frac{3}{2} \mathrm{E}_{1}$ $\mathrm{E}_{2}=\frac{3}{2} \times 3.2 \times 10^{-19}$ $\mathrm{E}_{2}=4.80 \times 10^{-19} \mathrm{~J}$
GUJCET 2018
Dual nature of radiation and Matter
142367
If $k_{\alpha}$ radiation of $M o(Z=42)$ has a wavelength $0.71 \AA$, then the wavelength of the corresponding radiation of $\mathrm{Cu}(\mathrm{Z}=29)$ is
1 $0.71 \AA$
2 $1.42 \AA$
3 $1.52 \AA$
4 $2.12 \AA$
Explanation:
C Given that, $\mathrm{Z}_{1}=42$ and $\mathrm{Z}_{2}=29, \lambda=0.71 \AA$ We know the relation - $\lambda \propto \frac{1}{(Z-b)^{2}}$ Then, $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(Z_{1}-b\right)^{2}}{\left(Z_{2}-b\right)^{2}}$ $\frac{\lambda_{2}}{0.71 \AA}=\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{(41)^{2}}{(28)^{2}}=\frac{1681}{784}=2.144$ So, $\quad \lambda_{2}=1.522 \AA$
CG PET- 2017
Dual nature of radiation and Matter
142370
A gamma ray photon creates an electron, positron pair. If the rest mass energy of an electron is $0.5 \mathrm{MeV}$ and the total $\mathrm{KE}$ of the electron, positron pair is $0.78 \mathrm{MeV}$, then the energy of the gamma ray photon must be
1 $0.78 \mathrm{MeV}$
2 $1.78 \mathrm{MeV}$
3 $1.28 \mathrm{MeV}$
4 $0.28 \mathrm{MeV}$
Explanation:
B Given that, Rest mass energy of electron $=0.5 \mathrm{MeV}$ Rest mass energy of positron $=0.5 \mathrm{MeV}$ $\because$ Mass energy of electron $=$ mass energy positron Total kinetic energy of the electron $=0.78 \mathrm{MeV}$ Energy of gamma rays photon $=(0.5+0.5+0.78)$ $=1.78 \mathrm{MeV}$
UPSEE - 2014
Dual nature of radiation and Matter
142371
The work function of sodium is $2.3 \mathrm{eV}$. The threshold wavelength of sodium will be
1 $2900 \AA$
2 $2500 \AA$
3 $5380 \AA$
4 $2000 \AA$
Explanation:
C Given that, Work function $(\phi)=2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \mathrm{~J}$ We know that, $\phi =\mathrm{h} v_{0}=\frac{\mathrm{hc}}{\lambda_{0}}$ $\text { Or } \quad \lambda_{0} =\frac{\mathrm{hc}}{\phi}$ $\lambda_{0} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.3 \times 1.6 \times 10^{-19}}$ $\lambda_{0}=\frac{19.8 \times 10^{-7}}{3.6}=5.396 \times 10^{-7} \mathrm{~m}$ $\lambda_{0}=5396 \AA \approx 5380 \AA$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142366
It the energy of photons corresponding to wavelength of $6000 \AA$ is $3.2 \times 10^{-19} \mathrm{~J}$. The photon energy for wavelength of $4000 \AA$ will be
1 $4.44 \times 10^{-19} \mathrm{~J}$
2 $2.22 \times 10^{-19} \mathrm{~J}$
3 $1.11 \times 10^{-19} \mathrm{~J}$
4 $4.80 \times 10^{-19} \mathrm{~J}$
Explanation:
D Given that, $\lambda_{1}=6000 \AA, \lambda_{2}=4000 \AA, \mathrm{E}_{1}=$ $3.2 \times 10^{-19} \mathrm{~J}$ We know that, $E=\frac{h c}{\lambda}$ $\text { So, } \quad E \propto \frac{1}{\lambda}$ Therefore, $\frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{4000}{6000}=\frac{2}{3}$ $\mathrm{E}_{2}=\frac{3}{2} \mathrm{E}_{1}$ $\mathrm{E}_{2}=\frac{3}{2} \times 3.2 \times 10^{-19}$ $\mathrm{E}_{2}=4.80 \times 10^{-19} \mathrm{~J}$
GUJCET 2018
Dual nature of radiation and Matter
142367
If $k_{\alpha}$ radiation of $M o(Z=42)$ has a wavelength $0.71 \AA$, then the wavelength of the corresponding radiation of $\mathrm{Cu}(\mathrm{Z}=29)$ is
1 $0.71 \AA$
2 $1.42 \AA$
3 $1.52 \AA$
4 $2.12 \AA$
Explanation:
C Given that, $\mathrm{Z}_{1}=42$ and $\mathrm{Z}_{2}=29, \lambda=0.71 \AA$ We know the relation - $\lambda \propto \frac{1}{(Z-b)^{2}}$ Then, $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(Z_{1}-b\right)^{2}}{\left(Z_{2}-b\right)^{2}}$ $\frac{\lambda_{2}}{0.71 \AA}=\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{(41)^{2}}{(28)^{2}}=\frac{1681}{784}=2.144$ So, $\quad \lambda_{2}=1.522 \AA$
CG PET- 2017
Dual nature of radiation and Matter
142370
A gamma ray photon creates an electron, positron pair. If the rest mass energy of an electron is $0.5 \mathrm{MeV}$ and the total $\mathrm{KE}$ of the electron, positron pair is $0.78 \mathrm{MeV}$, then the energy of the gamma ray photon must be
1 $0.78 \mathrm{MeV}$
2 $1.78 \mathrm{MeV}$
3 $1.28 \mathrm{MeV}$
4 $0.28 \mathrm{MeV}$
Explanation:
B Given that, Rest mass energy of electron $=0.5 \mathrm{MeV}$ Rest mass energy of positron $=0.5 \mathrm{MeV}$ $\because$ Mass energy of electron $=$ mass energy positron Total kinetic energy of the electron $=0.78 \mathrm{MeV}$ Energy of gamma rays photon $=(0.5+0.5+0.78)$ $=1.78 \mathrm{MeV}$
UPSEE - 2014
Dual nature of radiation and Matter
142371
The work function of sodium is $2.3 \mathrm{eV}$. The threshold wavelength of sodium will be
1 $2900 \AA$
2 $2500 \AA$
3 $5380 \AA$
4 $2000 \AA$
Explanation:
C Given that, Work function $(\phi)=2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \mathrm{~J}$ We know that, $\phi =\mathrm{h} v_{0}=\frac{\mathrm{hc}}{\lambda_{0}}$ $\text { Or } \quad \lambda_{0} =\frac{\mathrm{hc}}{\phi}$ $\lambda_{0} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.3 \times 1.6 \times 10^{-19}}$ $\lambda_{0}=\frac{19.8 \times 10^{-7}}{3.6}=5.396 \times 10^{-7} \mathrm{~m}$ $\lambda_{0}=5396 \AA \approx 5380 \AA$
142366
It the energy of photons corresponding to wavelength of $6000 \AA$ is $3.2 \times 10^{-19} \mathrm{~J}$. The photon energy for wavelength of $4000 \AA$ will be
1 $4.44 \times 10^{-19} \mathrm{~J}$
2 $2.22 \times 10^{-19} \mathrm{~J}$
3 $1.11 \times 10^{-19} \mathrm{~J}$
4 $4.80 \times 10^{-19} \mathrm{~J}$
Explanation:
D Given that, $\lambda_{1}=6000 \AA, \lambda_{2}=4000 \AA, \mathrm{E}_{1}=$ $3.2 \times 10^{-19} \mathrm{~J}$ We know that, $E=\frac{h c}{\lambda}$ $\text { So, } \quad E \propto \frac{1}{\lambda}$ Therefore, $\frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{4000}{6000}=\frac{2}{3}$ $\mathrm{E}_{2}=\frac{3}{2} \mathrm{E}_{1}$ $\mathrm{E}_{2}=\frac{3}{2} \times 3.2 \times 10^{-19}$ $\mathrm{E}_{2}=4.80 \times 10^{-19} \mathrm{~J}$
GUJCET 2018
Dual nature of radiation and Matter
142367
If $k_{\alpha}$ radiation of $M o(Z=42)$ has a wavelength $0.71 \AA$, then the wavelength of the corresponding radiation of $\mathrm{Cu}(\mathrm{Z}=29)$ is
1 $0.71 \AA$
2 $1.42 \AA$
3 $1.52 \AA$
4 $2.12 \AA$
Explanation:
C Given that, $\mathrm{Z}_{1}=42$ and $\mathrm{Z}_{2}=29, \lambda=0.71 \AA$ We know the relation - $\lambda \propto \frac{1}{(Z-b)^{2}}$ Then, $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(Z_{1}-b\right)^{2}}{\left(Z_{2}-b\right)^{2}}$ $\frac{\lambda_{2}}{0.71 \AA}=\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{(41)^{2}}{(28)^{2}}=\frac{1681}{784}=2.144$ So, $\quad \lambda_{2}=1.522 \AA$
CG PET- 2017
Dual nature of radiation and Matter
142370
A gamma ray photon creates an electron, positron pair. If the rest mass energy of an electron is $0.5 \mathrm{MeV}$ and the total $\mathrm{KE}$ of the electron, positron pair is $0.78 \mathrm{MeV}$, then the energy of the gamma ray photon must be
1 $0.78 \mathrm{MeV}$
2 $1.78 \mathrm{MeV}$
3 $1.28 \mathrm{MeV}$
4 $0.28 \mathrm{MeV}$
Explanation:
B Given that, Rest mass energy of electron $=0.5 \mathrm{MeV}$ Rest mass energy of positron $=0.5 \mathrm{MeV}$ $\because$ Mass energy of electron $=$ mass energy positron Total kinetic energy of the electron $=0.78 \mathrm{MeV}$ Energy of gamma rays photon $=(0.5+0.5+0.78)$ $=1.78 \mathrm{MeV}$
UPSEE - 2014
Dual nature of radiation and Matter
142371
The work function of sodium is $2.3 \mathrm{eV}$. The threshold wavelength of sodium will be
1 $2900 \AA$
2 $2500 \AA$
3 $5380 \AA$
4 $2000 \AA$
Explanation:
C Given that, Work function $(\phi)=2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \mathrm{~J}$ We know that, $\phi =\mathrm{h} v_{0}=\frac{\mathrm{hc}}{\lambda_{0}}$ $\text { Or } \quad \lambda_{0} =\frac{\mathrm{hc}}{\phi}$ $\lambda_{0} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.3 \times 1.6 \times 10^{-19}}$ $\lambda_{0}=\frac{19.8 \times 10^{-7}}{3.6}=5.396 \times 10^{-7} \mathrm{~m}$ $\lambda_{0}=5396 \AA \approx 5380 \AA$
142366
It the energy of photons corresponding to wavelength of $6000 \AA$ is $3.2 \times 10^{-19} \mathrm{~J}$. The photon energy for wavelength of $4000 \AA$ will be
1 $4.44 \times 10^{-19} \mathrm{~J}$
2 $2.22 \times 10^{-19} \mathrm{~J}$
3 $1.11 \times 10^{-19} \mathrm{~J}$
4 $4.80 \times 10^{-19} \mathrm{~J}$
Explanation:
D Given that, $\lambda_{1}=6000 \AA, \lambda_{2}=4000 \AA, \mathrm{E}_{1}=$ $3.2 \times 10^{-19} \mathrm{~J}$ We know that, $E=\frac{h c}{\lambda}$ $\text { So, } \quad E \propto \frac{1}{\lambda}$ Therefore, $\frac{E_{1}}{E_{2}}=\frac{\lambda_{2}}{\lambda_{1}}=\frac{4000}{6000}=\frac{2}{3}$ $\mathrm{E}_{2}=\frac{3}{2} \mathrm{E}_{1}$ $\mathrm{E}_{2}=\frac{3}{2} \times 3.2 \times 10^{-19}$ $\mathrm{E}_{2}=4.80 \times 10^{-19} \mathrm{~J}$
GUJCET 2018
Dual nature of radiation and Matter
142367
If $k_{\alpha}$ radiation of $M o(Z=42)$ has a wavelength $0.71 \AA$, then the wavelength of the corresponding radiation of $\mathrm{Cu}(\mathrm{Z}=29)$ is
1 $0.71 \AA$
2 $1.42 \AA$
3 $1.52 \AA$
4 $2.12 \AA$
Explanation:
C Given that, $\mathrm{Z}_{1}=42$ and $\mathrm{Z}_{2}=29, \lambda=0.71 \AA$ We know the relation - $\lambda \propto \frac{1}{(Z-b)^{2}}$ Then, $\frac{\lambda_{2}}{\lambda_{1}}=\frac{\left(Z_{1}-b\right)^{2}}{\left(Z_{2}-b\right)^{2}}$ $\frac{\lambda_{2}}{0.71 \AA}=\frac{(42-1)^{2}}{(29-1)^{2}}=\frac{(41)^{2}}{(28)^{2}}=\frac{1681}{784}=2.144$ So, $\quad \lambda_{2}=1.522 \AA$
CG PET- 2017
Dual nature of radiation and Matter
142370
A gamma ray photon creates an electron, positron pair. If the rest mass energy of an electron is $0.5 \mathrm{MeV}$ and the total $\mathrm{KE}$ of the electron, positron pair is $0.78 \mathrm{MeV}$, then the energy of the gamma ray photon must be
1 $0.78 \mathrm{MeV}$
2 $1.78 \mathrm{MeV}$
3 $1.28 \mathrm{MeV}$
4 $0.28 \mathrm{MeV}$
Explanation:
B Given that, Rest mass energy of electron $=0.5 \mathrm{MeV}$ Rest mass energy of positron $=0.5 \mathrm{MeV}$ $\because$ Mass energy of electron $=$ mass energy positron Total kinetic energy of the electron $=0.78 \mathrm{MeV}$ Energy of gamma rays photon $=(0.5+0.5+0.78)$ $=1.78 \mathrm{MeV}$
UPSEE - 2014
Dual nature of radiation and Matter
142371
The work function of sodium is $2.3 \mathrm{eV}$. The threshold wavelength of sodium will be
1 $2900 \AA$
2 $2500 \AA$
3 $5380 \AA$
4 $2000 \AA$
Explanation:
C Given that, Work function $(\phi)=2.3 \mathrm{eV}=2.3 \times 1.6 \times 10^{-19} \mathrm{~J}$ We know that, $\phi =\mathrm{h} v_{0}=\frac{\mathrm{hc}}{\lambda_{0}}$ $\text { Or } \quad \lambda_{0} =\frac{\mathrm{hc}}{\phi}$ $\lambda_{0} =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{2.3 \times 1.6 \times 10^{-19}}$ $\lambda_{0}=\frac{19.8 \times 10^{-7}}{3.6}=5.396 \times 10^{-7} \mathrm{~m}$ $\lambda_{0}=5396 \AA \approx 5380 \AA$