145740
The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \AA$ is
1 $602 \mathrm{~V}$
2 $50 \mathrm{~V}$
3 $138 \mathrm{~V}$
4 $812 \mathrm{~V}$
Explanation:
A Given, $\lambda=0.5 \AA, \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\left(\frac{\lambda}{\mathrm{h}}\right)=\frac{1}{\sqrt{2 \mathrm{meV}}}$ $\sqrt{2 \mathrm{meV}}=\frac{\mathrm{h}}{\lambda}$ Squaring on both sides, we get- $(\sqrt{2 \mathrm{meV}})^{2}=\left(\frac{\mathrm{h}}{\lambda}\right)^{2}$ $2 \mathrm{meV}=\frac{\mathrm{h}^{2}}{\lambda^{2}}$ $\mathrm{~V}=\frac{\mathrm{h}^{2}}{2 \mathrm{me} \lambda^{2}}$ $\mathrm{~V}=\frac{\left(6.62 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(5 \times 10^{-11}\right)^{2}}$ $\mathrm{~V}=602 \mathrm{~V}$
TSEAMCET 18.07.2022
ATOMS
145742
The period of revolution of an electron revolving in nth orbit of $\mathrm{H}$-atom is proportional to :
1 $\mathrm{n}^{2}$
2 $\frac{1}{n}$
3 $\mathrm{n}^{3}$
4 independent of $n$
Explanation:
C According to Bohr's model, $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ And $\frac{\mathrm{kZe}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ Time period of revolution of electron, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$ $\mathrm{T}=\frac{4 \varepsilon_{0}^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{mZ^{2 }} \mathrm{e}^{4}}$ $r \propto n^{2} / Z^{2} \text { and } v \propto Z / n$ For $\mathrm{H}$-atom $\mathrm{T} \propto \mathrm{n}^{3} / \mathrm{Z}^{2}$ For $\quad \mathrm{Z}=1$ $\therefore \quad \mathrm{T} \propto \mathrm{n}$ Thus, the time period is directly proportional to $\mathrm{n}^{3}$.
Karnataka CET-2020
ATOMS
145743
A hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \mathrm{~V}$. The speed of the electron when it strikes the anode is :
145744
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 $\mathrm{B}, \mathrm{C}$
2 C, D
3 $\mathrm{A}, \mathrm{C}$
4 A, D
Explanation:
D We know that the difference of energy between two level is equal to the energy of photon. $\Delta \mathrm{E}=\mathrm{h} v$ $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ The value of $\Delta \mathrm{E}$ is minimum for $\mathrm{A}$, So radiation corresponding to $\mathrm{A}$ have maximum wavelength and value of $\Delta \mathrm{E}$ is maximum for $\mathrm{D}$, So radiation corresponding to $\mathrm{D}$ have minimum wavelength. (i) For maximum wavelength energy should be minimum. (ii) For minimum wavelength energy should be maximum.
145740
The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \AA$ is
1 $602 \mathrm{~V}$
2 $50 \mathrm{~V}$
3 $138 \mathrm{~V}$
4 $812 \mathrm{~V}$
Explanation:
A Given, $\lambda=0.5 \AA, \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\left(\frac{\lambda}{\mathrm{h}}\right)=\frac{1}{\sqrt{2 \mathrm{meV}}}$ $\sqrt{2 \mathrm{meV}}=\frac{\mathrm{h}}{\lambda}$ Squaring on both sides, we get- $(\sqrt{2 \mathrm{meV}})^{2}=\left(\frac{\mathrm{h}}{\lambda}\right)^{2}$ $2 \mathrm{meV}=\frac{\mathrm{h}^{2}}{\lambda^{2}}$ $\mathrm{~V}=\frac{\mathrm{h}^{2}}{2 \mathrm{me} \lambda^{2}}$ $\mathrm{~V}=\frac{\left(6.62 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(5 \times 10^{-11}\right)^{2}}$ $\mathrm{~V}=602 \mathrm{~V}$
TSEAMCET 18.07.2022
ATOMS
145742
The period of revolution of an electron revolving in nth orbit of $\mathrm{H}$-atom is proportional to :
1 $\mathrm{n}^{2}$
2 $\frac{1}{n}$
3 $\mathrm{n}^{3}$
4 independent of $n$
Explanation:
C According to Bohr's model, $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ And $\frac{\mathrm{kZe}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ Time period of revolution of electron, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$ $\mathrm{T}=\frac{4 \varepsilon_{0}^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{mZ^{2 }} \mathrm{e}^{4}}$ $r \propto n^{2} / Z^{2} \text { and } v \propto Z / n$ For $\mathrm{H}$-atom $\mathrm{T} \propto \mathrm{n}^{3} / \mathrm{Z}^{2}$ For $\quad \mathrm{Z}=1$ $\therefore \quad \mathrm{T} \propto \mathrm{n}$ Thus, the time period is directly proportional to $\mathrm{n}^{3}$.
Karnataka CET-2020
ATOMS
145743
A hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \mathrm{~V}$. The speed of the electron when it strikes the anode is :
145744
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 $\mathrm{B}, \mathrm{C}$
2 C, D
3 $\mathrm{A}, \mathrm{C}$
4 A, D
Explanation:
D We know that the difference of energy between two level is equal to the energy of photon. $\Delta \mathrm{E}=\mathrm{h} v$ $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ The value of $\Delta \mathrm{E}$ is minimum for $\mathrm{A}$, So radiation corresponding to $\mathrm{A}$ have maximum wavelength and value of $\Delta \mathrm{E}$ is maximum for $\mathrm{D}$, So radiation corresponding to $\mathrm{D}$ have minimum wavelength. (i) For maximum wavelength energy should be minimum. (ii) For minimum wavelength energy should be maximum.
145740
The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \AA$ is
1 $602 \mathrm{~V}$
2 $50 \mathrm{~V}$
3 $138 \mathrm{~V}$
4 $812 \mathrm{~V}$
Explanation:
A Given, $\lambda=0.5 \AA, \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\left(\frac{\lambda}{\mathrm{h}}\right)=\frac{1}{\sqrt{2 \mathrm{meV}}}$ $\sqrt{2 \mathrm{meV}}=\frac{\mathrm{h}}{\lambda}$ Squaring on both sides, we get- $(\sqrt{2 \mathrm{meV}})^{2}=\left(\frac{\mathrm{h}}{\lambda}\right)^{2}$ $2 \mathrm{meV}=\frac{\mathrm{h}^{2}}{\lambda^{2}}$ $\mathrm{~V}=\frac{\mathrm{h}^{2}}{2 \mathrm{me} \lambda^{2}}$ $\mathrm{~V}=\frac{\left(6.62 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(5 \times 10^{-11}\right)^{2}}$ $\mathrm{~V}=602 \mathrm{~V}$
TSEAMCET 18.07.2022
ATOMS
145742
The period of revolution of an electron revolving in nth orbit of $\mathrm{H}$-atom is proportional to :
1 $\mathrm{n}^{2}$
2 $\frac{1}{n}$
3 $\mathrm{n}^{3}$
4 independent of $n$
Explanation:
C According to Bohr's model, $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ And $\frac{\mathrm{kZe}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ Time period of revolution of electron, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$ $\mathrm{T}=\frac{4 \varepsilon_{0}^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{mZ^{2 }} \mathrm{e}^{4}}$ $r \propto n^{2} / Z^{2} \text { and } v \propto Z / n$ For $\mathrm{H}$-atom $\mathrm{T} \propto \mathrm{n}^{3} / \mathrm{Z}^{2}$ For $\quad \mathrm{Z}=1$ $\therefore \quad \mathrm{T} \propto \mathrm{n}$ Thus, the time period is directly proportional to $\mathrm{n}^{3}$.
Karnataka CET-2020
ATOMS
145743
A hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \mathrm{~V}$. The speed of the electron when it strikes the anode is :
145744
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 $\mathrm{B}, \mathrm{C}$
2 C, D
3 $\mathrm{A}, \mathrm{C}$
4 A, D
Explanation:
D We know that the difference of energy between two level is equal to the energy of photon. $\Delta \mathrm{E}=\mathrm{h} v$ $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ The value of $\Delta \mathrm{E}$ is minimum for $\mathrm{A}$, So radiation corresponding to $\mathrm{A}$ have maximum wavelength and value of $\Delta \mathrm{E}$ is maximum for $\mathrm{D}$, So radiation corresponding to $\mathrm{D}$ have minimum wavelength. (i) For maximum wavelength energy should be minimum. (ii) For minimum wavelength energy should be maximum.
145740
The voltage applied to an electron microscope to produce electrons of wavelength $0.50 \AA$ is
1 $602 \mathrm{~V}$
2 $50 \mathrm{~V}$
3 $138 \mathrm{~V}$
4 $812 \mathrm{~V}$
Explanation:
A Given, $\lambda=0.5 \AA, \mathrm{m}=9.1 \times 10^{-31} \mathrm{~kg}$ We know that, $\lambda=\frac{\mathrm{h}}{\sqrt{2 \mathrm{meV}}}$ $\left(\frac{\lambda}{\mathrm{h}}\right)=\frac{1}{\sqrt{2 \mathrm{meV}}}$ $\sqrt{2 \mathrm{meV}}=\frac{\mathrm{h}}{\lambda}$ Squaring on both sides, we get- $(\sqrt{2 \mathrm{meV}})^{2}=\left(\frac{\mathrm{h}}{\lambda}\right)^{2}$ $2 \mathrm{meV}=\frac{\mathrm{h}^{2}}{\lambda^{2}}$ $\mathrm{~V}=\frac{\mathrm{h}^{2}}{2 \mathrm{me} \lambda^{2}}$ $\mathrm{~V}=\frac{\left(6.62 \times 10^{-34}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19} \times\left(5 \times 10^{-11}\right)^{2}}$ $\mathrm{~V}=602 \mathrm{~V}$
TSEAMCET 18.07.2022
ATOMS
145742
The period of revolution of an electron revolving in nth orbit of $\mathrm{H}$-atom is proportional to :
1 $\mathrm{n}^{2}$
2 $\frac{1}{n}$
3 $\mathrm{n}^{3}$
4 independent of $n$
Explanation:
C According to Bohr's model, $\mathrm{mvr}=\frac{\mathrm{nh}}{2 \pi}$ And $\frac{\mathrm{kZe}^{2}}{\mathrm{r}^{2}}=\frac{\mathrm{mv}^{2}}{\mathrm{r}}$ Time period of revolution of electron, $\mathrm{T}=\frac{2 \pi \mathrm{r}}{\mathrm{V}}$ $\mathrm{T}=\frac{4 \varepsilon_{0}^{2} \mathrm{n}^{3} \mathrm{~h}^{3}}{\mathrm{mZ^{2 }} \mathrm{e}^{4}}$ $r \propto n^{2} / Z^{2} \text { and } v \propto Z / n$ For $\mathrm{H}$-atom $\mathrm{T} \propto \mathrm{n}^{3} / \mathrm{Z}^{2}$ For $\quad \mathrm{Z}=1$ $\therefore \quad \mathrm{T} \propto \mathrm{n}$ Thus, the time period is directly proportional to $\mathrm{n}^{3}$.
Karnataka CET-2020
ATOMS
145743
A hot filament liberates an electron with zero initial velocity. The anode potential is $1200 \mathrm{~V}$. The speed of the electron when it strikes the anode is :
145744
The energy levels with transitions for the atom are shown. The transitions corresponding to emission of radiation of maximum and minimum wavelength are respectively
1 $\mathrm{B}, \mathrm{C}$
2 C, D
3 $\mathrm{A}, \mathrm{C}$
4 A, D
Explanation:
D We know that the difference of energy between two level is equal to the energy of photon. $\Delta \mathrm{E}=\mathrm{h} v$ $\Delta \mathrm{E}=\frac{\mathrm{hc}}{\lambda}$ $\lambda=\frac{\mathrm{hc}}{\Delta \mathrm{E}}$ $\lambda \propto \frac{1}{\Delta \mathrm{E}}$ The value of $\Delta \mathrm{E}$ is minimum for $\mathrm{A}$, So radiation corresponding to $\mathrm{A}$ have maximum wavelength and value of $\Delta \mathrm{E}$ is maximum for $\mathrm{D}$, So radiation corresponding to $\mathrm{D}$ have minimum wavelength. (i) For maximum wavelength energy should be minimum. (ii) For minimum wavelength energy should be maximum.
