145253
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B Both electron and photon propagate in the form of waves with same wavelength, it means that they have same momentum. We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ If $\lambda$ is same, then momentum (p) will be same.
Manipal UGET-2010
ATOMS
145262
$\mathrm{X}$ - rays are produced by accelerating electrons by voltage $V$ and let they strike a metal of atomic number $\mathrm{Z}$. The highest frequency of $\mathrm{X}$ rays produced is proportional to :
1 $\mathrm{V}$
2 $\mathrm{Z}$
3 $(\mathrm{Z}-1)$
4 $(Z-1)^{2}$
Explanation:
D According to mosleys law, $v=\mathrm{a}(\mathrm{Z}-\mathrm{b})^{2}$ Where, $\mathrm{a}$ and $\mathrm{b}$ are constant. Neglecting the effect of outer electros and other L electrons, the electrons making the transition finds a charge $\mathrm{a}(\mathrm{Z}-1)^{2}$, putting $\mathrm{b}=1$. $v_{\max }=\mathrm{a}(\mathrm{Z}-1)^{2}$ $v_{\max } \propto(\mathrm{Z}-1)^{2}$
UPSEE - 2004
ATOMS
145276
The ionization energy of 10 times ionized sodium atom is :
1 $\frac{13.6}{11} \mathrm{eV}$
2 $\frac{13.6}{112} \mathrm{eV}$
3 $13.6 \times(11)^{2} \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C The energy of $\mathrm{n}^{\text {th }}$ orbit of hydrogen like atom in, $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$ Here, $Z=11$ and number of 10 electrons are removed already. For the last electron to be removed $n=1$ $\therefore \mathrm{E}_{\mathrm{n}}=\frac{-13.6 \times(11)^{2}}{(1)^{2}} \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=-13.6 \times(11)^{2} \mathrm{eV}$
BCECE-2006
ATOMS
145288
If $M$ is the atomic mass and $A$ is the mass number, packing fraction is given by :
1 $\frac{\mathrm{M}}{\mathrm{M}-\mathrm{A}}$
2 $\frac{M-A}{A}$
3 $\frac{\mathrm{A}}{\mathrm{M}-\mathrm{A}}$
4 $\frac{\mathrm{A}-\mathrm{M}}{\mathrm{A}}$
Explanation:
B Given, Atomic mass $=\mathrm{M}$ Mass number $=\mathrm{A}$ Packing fraction is equal to the difference of atomic mass and mass number whole divided by the mass number Packing fraction $=\frac{\mathrm{M}-\mathrm{A}}{\mathrm{A}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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ATOMS
145253
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B Both electron and photon propagate in the form of waves with same wavelength, it means that they have same momentum. We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ If $\lambda$ is same, then momentum (p) will be same.
Manipal UGET-2010
ATOMS
145262
$\mathrm{X}$ - rays are produced by accelerating electrons by voltage $V$ and let they strike a metal of atomic number $\mathrm{Z}$. The highest frequency of $\mathrm{X}$ rays produced is proportional to :
1 $\mathrm{V}$
2 $\mathrm{Z}$
3 $(\mathrm{Z}-1)$
4 $(Z-1)^{2}$
Explanation:
D According to mosleys law, $v=\mathrm{a}(\mathrm{Z}-\mathrm{b})^{2}$ Where, $\mathrm{a}$ and $\mathrm{b}$ are constant. Neglecting the effect of outer electros and other L electrons, the electrons making the transition finds a charge $\mathrm{a}(\mathrm{Z}-1)^{2}$, putting $\mathrm{b}=1$. $v_{\max }=\mathrm{a}(\mathrm{Z}-1)^{2}$ $v_{\max } \propto(\mathrm{Z}-1)^{2}$
UPSEE - 2004
ATOMS
145276
The ionization energy of 10 times ionized sodium atom is :
1 $\frac{13.6}{11} \mathrm{eV}$
2 $\frac{13.6}{112} \mathrm{eV}$
3 $13.6 \times(11)^{2} \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C The energy of $\mathrm{n}^{\text {th }}$ orbit of hydrogen like atom in, $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$ Here, $Z=11$ and number of 10 electrons are removed already. For the last electron to be removed $n=1$ $\therefore \mathrm{E}_{\mathrm{n}}=\frac{-13.6 \times(11)^{2}}{(1)^{2}} \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=-13.6 \times(11)^{2} \mathrm{eV}$
BCECE-2006
ATOMS
145288
If $M$ is the atomic mass and $A$ is the mass number, packing fraction is given by :
1 $\frac{\mathrm{M}}{\mathrm{M}-\mathrm{A}}$
2 $\frac{M-A}{A}$
3 $\frac{\mathrm{A}}{\mathrm{M}-\mathrm{A}}$
4 $\frac{\mathrm{A}-\mathrm{M}}{\mathrm{A}}$
Explanation:
B Given, Atomic mass $=\mathrm{M}$ Mass number $=\mathrm{A}$ Packing fraction is equal to the difference of atomic mass and mass number whole divided by the mass number Packing fraction $=\frac{\mathrm{M}-\mathrm{A}}{\mathrm{A}}$
145253
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B Both electron and photon propagate in the form of waves with same wavelength, it means that they have same momentum. We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ If $\lambda$ is same, then momentum (p) will be same.
Manipal UGET-2010
ATOMS
145262
$\mathrm{X}$ - rays are produced by accelerating electrons by voltage $V$ and let they strike a metal of atomic number $\mathrm{Z}$. The highest frequency of $\mathrm{X}$ rays produced is proportional to :
1 $\mathrm{V}$
2 $\mathrm{Z}$
3 $(\mathrm{Z}-1)$
4 $(Z-1)^{2}$
Explanation:
D According to mosleys law, $v=\mathrm{a}(\mathrm{Z}-\mathrm{b})^{2}$ Where, $\mathrm{a}$ and $\mathrm{b}$ are constant. Neglecting the effect of outer electros and other L electrons, the electrons making the transition finds a charge $\mathrm{a}(\mathrm{Z}-1)^{2}$, putting $\mathrm{b}=1$. $v_{\max }=\mathrm{a}(\mathrm{Z}-1)^{2}$ $v_{\max } \propto(\mathrm{Z}-1)^{2}$
UPSEE - 2004
ATOMS
145276
The ionization energy of 10 times ionized sodium atom is :
1 $\frac{13.6}{11} \mathrm{eV}$
2 $\frac{13.6}{112} \mathrm{eV}$
3 $13.6 \times(11)^{2} \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C The energy of $\mathrm{n}^{\text {th }}$ orbit of hydrogen like atom in, $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$ Here, $Z=11$ and number of 10 electrons are removed already. For the last electron to be removed $n=1$ $\therefore \mathrm{E}_{\mathrm{n}}=\frac{-13.6 \times(11)^{2}}{(1)^{2}} \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=-13.6 \times(11)^{2} \mathrm{eV}$
BCECE-2006
ATOMS
145288
If $M$ is the atomic mass and $A$ is the mass number, packing fraction is given by :
1 $\frac{\mathrm{M}}{\mathrm{M}-\mathrm{A}}$
2 $\frac{M-A}{A}$
3 $\frac{\mathrm{A}}{\mathrm{M}-\mathrm{A}}$
4 $\frac{\mathrm{A}-\mathrm{M}}{\mathrm{A}}$
Explanation:
B Given, Atomic mass $=\mathrm{M}$ Mass number $=\mathrm{A}$ Packing fraction is equal to the difference of atomic mass and mass number whole divided by the mass number Packing fraction $=\frac{\mathrm{M}-\mathrm{A}}{\mathrm{A}}$
145253
If an electron and a photon propagate in the form of waves having the same wavelength, it implies that they have the same
1 energy
2 momentum
3 velocity
4 angular momentum
Explanation:
B Both electron and photon propagate in the form of waves with same wavelength, it means that they have same momentum. We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ If $\lambda$ is same, then momentum (p) will be same.
Manipal UGET-2010
ATOMS
145262
$\mathrm{X}$ - rays are produced by accelerating electrons by voltage $V$ and let they strike a metal of atomic number $\mathrm{Z}$. The highest frequency of $\mathrm{X}$ rays produced is proportional to :
1 $\mathrm{V}$
2 $\mathrm{Z}$
3 $(\mathrm{Z}-1)$
4 $(Z-1)^{2}$
Explanation:
D According to mosleys law, $v=\mathrm{a}(\mathrm{Z}-\mathrm{b})^{2}$ Where, $\mathrm{a}$ and $\mathrm{b}$ are constant. Neglecting the effect of outer electros and other L electrons, the electrons making the transition finds a charge $\mathrm{a}(\mathrm{Z}-1)^{2}$, putting $\mathrm{b}=1$. $v_{\max }=\mathrm{a}(\mathrm{Z}-1)^{2}$ $v_{\max } \propto(\mathrm{Z}-1)^{2}$
UPSEE - 2004
ATOMS
145276
The ionization energy of 10 times ionized sodium atom is :
1 $\frac{13.6}{11} \mathrm{eV}$
2 $\frac{13.6}{112} \mathrm{eV}$
3 $13.6 \times(11)^{2} \mathrm{eV}$
4 $13.6 \mathrm{eV}$
Explanation:
C The energy of $\mathrm{n}^{\text {th }}$ orbit of hydrogen like atom in, $E_{n}=-13.6 \frac{Z^{2}}{n^{2}}$ Here, $Z=11$ and number of 10 electrons are removed already. For the last electron to be removed $n=1$ $\therefore \mathrm{E}_{\mathrm{n}}=\frac{-13.6 \times(11)^{2}}{(1)^{2}} \mathrm{eV}$ $\mathrm{E}_{\mathrm{n}}=-13.6 \times(11)^{2} \mathrm{eV}$
BCECE-2006
ATOMS
145288
If $M$ is the atomic mass and $A$ is the mass number, packing fraction is given by :
1 $\frac{\mathrm{M}}{\mathrm{M}-\mathrm{A}}$
2 $\frac{M-A}{A}$
3 $\frac{\mathrm{A}}{\mathrm{M}-\mathrm{A}}$
4 $\frac{\mathrm{A}-\mathrm{M}}{\mathrm{A}}$
Explanation:
B Given, Atomic mass $=\mathrm{M}$ Mass number $=\mathrm{A}$ Packing fraction is equal to the difference of atomic mass and mass number whole divided by the mass number Packing fraction $=\frac{\mathrm{M}-\mathrm{A}}{\mathrm{A}}$