145761
The distance of closest approach for an alpha nucleus of velocity $v$ bombarding a stationary heavy nucleus target of charge $\mathrm{Ze}$ is directly proportional to
1 $\mathrm{V}$
2 $\mathrm{m}$
3 $\frac{1}{\mathrm{v}^{2}}$
4 $\frac{1}{\mathrm{Ze}}$
Explanation:
C Let the distance of closest approached Kinetic energy of $\alpha$ particle $=$ Potential energy of $\alpha$ particle $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{e}) \cdot \mathrm{Ze}}{\mathrm{d}}$ $\mathrm{d}=\frac{4 \mathrm{Ze}^{2}}{4 \pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d}=\frac{\mathrm{Ze} \mathrm{e}^{2}}{\pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d} \propto \frac{1}{\mathrm{v}^{2}}$
J and K CET-2016
ATOMS
145762
If the electron in a hydrogen atom jumps from an orbit with level $n_{1}=2$ to an orbit with level $\mathbf{n}_{2}=1$ the emitted radiation has a wavelength given by
1 $\lambda=\frac{5}{3 \mathrm{R}}$
2 $\lambda=\frac{4}{3 \mathrm{R}}$
3 $\lambda=\frac{\mathrm{R}}{4}$
4 $\lambda=\frac{3 \mathrm{R}}{4}$
Explanation:
B Given that, $\mathrm{n}_{1}=2$, and $\mathrm{n}_{1}=1$ We know that, Wavelength of $\mathrm{H}$ atom $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{4}$ $\lambda=\frac{4}{3 \mathrm{R}}$
WB JEE 2009
ATOMS
145763
Energy corresponding to a photon of wavelength $5700 \AA$ is
1 $21.7 \mathrm{eV}$
2 $2.17 \mathrm{eV}$
3 $8.34 \mathrm{eV}$
4 $16.68 \mathrm{eV}$
Explanation:
B Given that, $\mathrm{h}=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\lambda=5700 \AA=5700 \times 10^{-10} \mathrm{~m}$ We know that, Energy of photon $(E)=\frac{h c}{\lambda}$ $\mathrm{E}=\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{5700 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=2.17 \mathrm{eV}$
WB JEE-2007
ATOMS
145764
The wavelength of $K_{a} X$-rays for lead isotopes $\mathrm{Pb}^{208}, \mathbf{P b}{ }^{206}$ are $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ respectively, Then,
1 $\lambda_{1}=\lambda_{2}=\lambda_{3}$
2 $\lambda_{1}>\lambda_{2}>\lambda_{3}$
3 $\lambda_{1} \lt \lambda_{2} \lt \lambda_{3}$
4 $\lambda_{1}=\lambda_{2}>\lambda_{3}$
Explanation:
A Wavelengths of $\mathrm{k}_{\alpha}$ lines for given isotopes $\frac{1}{\lambda}=\mathrm{R}(\mathrm{Z}-1)^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$ $\mathrm{Z}=82$ Where, $\mathrm{R}=$ Rydberg constant $\text { Then, } \frac{1}{\lambda_{1}}=\mathrm{R}(82-1)^{2}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}[81]^{2}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}(81)^{2}}{4}$ Similarly, $\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{3}}=\frac{3 \mathrm{R}(81)^{2}}{4} \quad \text { (because } \mathrm{Z} \text { are same) }$ Then $\lambda_{1}=\lambda_{2}=\lambda_{3}$
145761
The distance of closest approach for an alpha nucleus of velocity $v$ bombarding a stationary heavy nucleus target of charge $\mathrm{Ze}$ is directly proportional to
1 $\mathrm{V}$
2 $\mathrm{m}$
3 $\frac{1}{\mathrm{v}^{2}}$
4 $\frac{1}{\mathrm{Ze}}$
Explanation:
C Let the distance of closest approached Kinetic energy of $\alpha$ particle $=$ Potential energy of $\alpha$ particle $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{e}) \cdot \mathrm{Ze}}{\mathrm{d}}$ $\mathrm{d}=\frac{4 \mathrm{Ze}^{2}}{4 \pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d}=\frac{\mathrm{Ze} \mathrm{e}^{2}}{\pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d} \propto \frac{1}{\mathrm{v}^{2}}$
J and K CET-2016
ATOMS
145762
If the electron in a hydrogen atom jumps from an orbit with level $n_{1}=2$ to an orbit with level $\mathbf{n}_{2}=1$ the emitted radiation has a wavelength given by
1 $\lambda=\frac{5}{3 \mathrm{R}}$
2 $\lambda=\frac{4}{3 \mathrm{R}}$
3 $\lambda=\frac{\mathrm{R}}{4}$
4 $\lambda=\frac{3 \mathrm{R}}{4}$
Explanation:
B Given that, $\mathrm{n}_{1}=2$, and $\mathrm{n}_{1}=1$ We know that, Wavelength of $\mathrm{H}$ atom $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{4}$ $\lambda=\frac{4}{3 \mathrm{R}}$
WB JEE 2009
ATOMS
145763
Energy corresponding to a photon of wavelength $5700 \AA$ is
1 $21.7 \mathrm{eV}$
2 $2.17 \mathrm{eV}$
3 $8.34 \mathrm{eV}$
4 $16.68 \mathrm{eV}$
Explanation:
B Given that, $\mathrm{h}=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\lambda=5700 \AA=5700 \times 10^{-10} \mathrm{~m}$ We know that, Energy of photon $(E)=\frac{h c}{\lambda}$ $\mathrm{E}=\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{5700 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=2.17 \mathrm{eV}$
WB JEE-2007
ATOMS
145764
The wavelength of $K_{a} X$-rays for lead isotopes $\mathrm{Pb}^{208}, \mathbf{P b}{ }^{206}$ are $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ respectively, Then,
1 $\lambda_{1}=\lambda_{2}=\lambda_{3}$
2 $\lambda_{1}>\lambda_{2}>\lambda_{3}$
3 $\lambda_{1} \lt \lambda_{2} \lt \lambda_{3}$
4 $\lambda_{1}=\lambda_{2}>\lambda_{3}$
Explanation:
A Wavelengths of $\mathrm{k}_{\alpha}$ lines for given isotopes $\frac{1}{\lambda}=\mathrm{R}(\mathrm{Z}-1)^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$ $\mathrm{Z}=82$ Where, $\mathrm{R}=$ Rydberg constant $\text { Then, } \frac{1}{\lambda_{1}}=\mathrm{R}(82-1)^{2}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}[81]^{2}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}(81)^{2}}{4}$ Similarly, $\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{3}}=\frac{3 \mathrm{R}(81)^{2}}{4} \quad \text { (because } \mathrm{Z} \text { are same) }$ Then $\lambda_{1}=\lambda_{2}=\lambda_{3}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
ATOMS
145761
The distance of closest approach for an alpha nucleus of velocity $v$ bombarding a stationary heavy nucleus target of charge $\mathrm{Ze}$ is directly proportional to
1 $\mathrm{V}$
2 $\mathrm{m}$
3 $\frac{1}{\mathrm{v}^{2}}$
4 $\frac{1}{\mathrm{Ze}}$
Explanation:
C Let the distance of closest approached Kinetic energy of $\alpha$ particle $=$ Potential energy of $\alpha$ particle $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{e}) \cdot \mathrm{Ze}}{\mathrm{d}}$ $\mathrm{d}=\frac{4 \mathrm{Ze}^{2}}{4 \pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d}=\frac{\mathrm{Ze} \mathrm{e}^{2}}{\pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d} \propto \frac{1}{\mathrm{v}^{2}}$
J and K CET-2016
ATOMS
145762
If the electron in a hydrogen atom jumps from an orbit with level $n_{1}=2$ to an orbit with level $\mathbf{n}_{2}=1$ the emitted radiation has a wavelength given by
1 $\lambda=\frac{5}{3 \mathrm{R}}$
2 $\lambda=\frac{4}{3 \mathrm{R}}$
3 $\lambda=\frac{\mathrm{R}}{4}$
4 $\lambda=\frac{3 \mathrm{R}}{4}$
Explanation:
B Given that, $\mathrm{n}_{1}=2$, and $\mathrm{n}_{1}=1$ We know that, Wavelength of $\mathrm{H}$ atom $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{4}$ $\lambda=\frac{4}{3 \mathrm{R}}$
WB JEE 2009
ATOMS
145763
Energy corresponding to a photon of wavelength $5700 \AA$ is
1 $21.7 \mathrm{eV}$
2 $2.17 \mathrm{eV}$
3 $8.34 \mathrm{eV}$
4 $16.68 \mathrm{eV}$
Explanation:
B Given that, $\mathrm{h}=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\lambda=5700 \AA=5700 \times 10^{-10} \mathrm{~m}$ We know that, Energy of photon $(E)=\frac{h c}{\lambda}$ $\mathrm{E}=\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{5700 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=2.17 \mathrm{eV}$
WB JEE-2007
ATOMS
145764
The wavelength of $K_{a} X$-rays for lead isotopes $\mathrm{Pb}^{208}, \mathbf{P b}{ }^{206}$ are $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ respectively, Then,
1 $\lambda_{1}=\lambda_{2}=\lambda_{3}$
2 $\lambda_{1}>\lambda_{2}>\lambda_{3}$
3 $\lambda_{1} \lt \lambda_{2} \lt \lambda_{3}$
4 $\lambda_{1}=\lambda_{2}>\lambda_{3}$
Explanation:
A Wavelengths of $\mathrm{k}_{\alpha}$ lines for given isotopes $\frac{1}{\lambda}=\mathrm{R}(\mathrm{Z}-1)^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$ $\mathrm{Z}=82$ Where, $\mathrm{R}=$ Rydberg constant $\text { Then, } \frac{1}{\lambda_{1}}=\mathrm{R}(82-1)^{2}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}[81]^{2}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}(81)^{2}}{4}$ Similarly, $\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{3}}=\frac{3 \mathrm{R}(81)^{2}}{4} \quad \text { (because } \mathrm{Z} \text { are same) }$ Then $\lambda_{1}=\lambda_{2}=\lambda_{3}$
145761
The distance of closest approach for an alpha nucleus of velocity $v$ bombarding a stationary heavy nucleus target of charge $\mathrm{Ze}$ is directly proportional to
1 $\mathrm{V}$
2 $\mathrm{m}$
3 $\frac{1}{\mathrm{v}^{2}}$
4 $\frac{1}{\mathrm{Ze}}$
Explanation:
C Let the distance of closest approached Kinetic energy of $\alpha$ particle $=$ Potential energy of $\alpha$ particle $\frac{1}{2} \mathrm{mv}^{2}=\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 \mathrm{e}) \cdot \mathrm{Ze}}{\mathrm{d}}$ $\mathrm{d}=\frac{4 \mathrm{Ze}^{2}}{4 \pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d}=\frac{\mathrm{Ze} \mathrm{e}^{2}}{\pi \varepsilon_{0} \mathrm{mv}^{2}}$ $\mathrm{~d} \propto \frac{1}{\mathrm{v}^{2}}$
J and K CET-2016
ATOMS
145762
If the electron in a hydrogen atom jumps from an orbit with level $n_{1}=2$ to an orbit with level $\mathbf{n}_{2}=1$ the emitted radiation has a wavelength given by
1 $\lambda=\frac{5}{3 \mathrm{R}}$
2 $\lambda=\frac{4}{3 \mathrm{R}}$
3 $\lambda=\frac{\mathrm{R}}{4}$
4 $\lambda=\frac{3 \mathrm{R}}{4}$
Explanation:
B Given that, $\mathrm{n}_{1}=2$, and $\mathrm{n}_{1}=1$ We know that, Wavelength of $\mathrm{H}$ atom $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{\mathrm{n}_{1}^{2}}-\frac{1}{\mathrm{n}_{2}^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{(1)^{2}}-\frac{1}{(2)^{2}}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda}=\mathrm{R}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda}=\frac{3 \mathrm{R}}{4}$ $\lambda=\frac{4}{3 \mathrm{R}}$
WB JEE 2009
ATOMS
145763
Energy corresponding to a photon of wavelength $5700 \AA$ is
1 $21.7 \mathrm{eV}$
2 $2.17 \mathrm{eV}$
3 $8.34 \mathrm{eV}$
4 $16.68 \mathrm{eV}$
Explanation:
B Given that, $\mathrm{h}=6.67 \times 10^{-34} \mathrm{~J}-\mathrm{s}$ $\mathrm{c}=3 \times 10^{8} \mathrm{~m} / \mathrm{s}$ $\lambda=5700 \AA=5700 \times 10^{-10} \mathrm{~m}$ We know that, Energy of photon $(E)=\frac{h c}{\lambda}$ $\mathrm{E}=\frac{6.67 \times 10^{-34} \times 3 \times 10^{8}}{5700 \times 10^{-10} \times 1.6 \times 10^{-19}} \mathrm{eV}$ $\mathrm{E}=2.17 \mathrm{eV}$
WB JEE-2007
ATOMS
145764
The wavelength of $K_{a} X$-rays for lead isotopes $\mathrm{Pb}^{208}, \mathbf{P b}{ }^{206}$ are $\lambda_{1}, \lambda_{2}$ and $\lambda_{3}$ respectively, Then,
1 $\lambda_{1}=\lambda_{2}=\lambda_{3}$
2 $\lambda_{1}>\lambda_{2}>\lambda_{3}$
3 $\lambda_{1} \lt \lambda_{2} \lt \lambda_{3}$
4 $\lambda_{1}=\lambda_{2}>\lambda_{3}$
Explanation:
A Wavelengths of $\mathrm{k}_{\alpha}$ lines for given isotopes $\frac{1}{\lambda}=\mathrm{R}(\mathrm{Z}-1)^{2}\left[\frac{1}{1^{2}}-\frac{1}{2^{2}}\right]$ $\mathrm{Z}=82$ Where, $\mathrm{R}=$ Rydberg constant $\text { Then, } \frac{1}{\lambda_{1}}=\mathrm{R}(82-1)^{2}\left[\frac{1}{1}-\frac{1}{4}\right]$ $\frac{1}{\lambda_{1}}=\mathrm{R}[81]^{2}\left[\frac{4-1}{4}\right]$ $\frac{1}{\lambda_{1}}=\frac{3 \mathrm{R}(81)^{2}}{4}$ Similarly, $\frac{1}{\lambda_{2}}=\frac{1}{\lambda_{3}}=\frac{3 \mathrm{R}(81)^{2}}{4} \quad \text { (because } \mathrm{Z} \text { are same) }$ Then $\lambda_{1}=\lambda_{2}=\lambda_{3}$