238852
The radial probability distribution curve obtained for an orbital wave function $(\psi)$ has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function $(\psi)$ correspond to ?
1 $\mathrm{Cu}$
2 $\mathrm{Li}$
3 $\mathrm{K}$
4 $\mathrm{Na}$
Explanation:
The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes i.e. $\mathrm{n}=3$ and $\mathrm{l}=0$ ( for radial node $=2$ ) Hence, the valence electron of the metal must present in $3 \mathrm{~s}-\mathrm{orbital}$. The electronic configuration of the given metal is - $\begin{aligned} & { }_{29} \mathrm{Cu}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^1 \\ & { }_3 \mathrm{Li}=[\mathrm{He}] 2 \mathrm{~s}^1 \\ & { }_{19}^{\mathrm{K}}=[\mathrm{Ar}] 4 \mathrm{~s}^1 \\ & { }_{11} \mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1 \end{aligned}$ Thus, the given wave function corresponds to the valence electrons of the sodium metal.
AP- EAMCET(Medical) -2010
Structure of Atom
238854
Assertion (A) : The probability of finding an electron in a small volume around a point $(x, y$, $z$ ) at a distance ' $r$ ' from the nucleus is proportional to $\psi^2$. Reason (R) : Subatomic particles have both wave and particle nature. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $(\mathrm{A})$
2 Both (A) and (R) are true but (R) is not the correct explanation of (A)
3 Only (A) is true but (R) is not true
4 (A) is not true but (R) is true
Explanation:
The square of the wave function, $\psi^2$ is called probability density and is always positive. The probability of finding an electron at a distance ' $r$ ' from the nucleus is expressed by following relation- Probability $=4 \pi \mathrm{r}^2 \mathrm{dr} \cdot \psi^2$ $\therefore \quad$ Probability $\propto \psi^2$ The de-Broglie explains any particle shows the both nature i.e. particle and wave nature. Both (A) and (R) are true but (R) is not the correct explanation of (A).
AP - EAMCET(MEDICAL) - 2009
Structure of Atom
238855
Wave number of spectral line for a given transition is $\mathrm{x} \mathrm{cm}^{-1}$ for $\mathrm{He}^{+}$, then its value for $\mathrm{Be}^{3+}$ (isoelectronic of $\mathrm{He}^{+}$) for same transition is-
1 $\frac{x}{4} \mathrm{~cm}^{-1}$
2 $\mathrm{x} \mathrm{cm}^{-1}$
3 $4 \mathrm{x} \mathrm{cm}-1$
4 $16 \mathrm{x} \mathrm{cm}^{-1}$
Explanation:
The spectral lines of the transitions are directly proportional to square of atomic number i.e. $\mathrm{z}^2$. $\bar{v} \propto z^2$ Given, $\quad \bar{v}_1=\mathrm{x}$ Now, $\quad \frac{\bar{v}_1}{\bar{v}_2}=\frac{\mathrm{z}_1^2}{\mathrm{z}_2^2}$ For, $\mathrm{He}^{+}, \mathrm{z}_1=2$ For, $\quad \mathrm{Be}^{3+}, \mathrm{z}_2=4$ $\begin{aligned} & \frac{\mathrm{x}}{\bar{v}_2}=\frac{(4)^2}{(2)^2} \\ & \bar{v}_2=4 \mathrm{x} \mathrm{cm}^{-1} \end{aligned}$
BCECE-2013
Structure of Atom
238857
The graph between $|\psi|^2$ and $r$ (radial distance) is shown below. This represents.
1 1 s-orbital
2 2 p-orbital
3 $3 \mathrm{~s}$-orbital
4 $2 \mathrm{~s}$-orbital
Explanation:
The given probability density curve is for $2 \mathrm{~s}$ orbital because it has only one radial node. Among other given orbitals, $1 \mathrm{~s}$ and $2 \mathrm{p}$ do not have any radial node and $3 \mathrm{~s}$ has two radial nodes.
238852
The radial probability distribution curve obtained for an orbital wave function $(\psi)$ has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function $(\psi)$ correspond to ?
1 $\mathrm{Cu}$
2 $\mathrm{Li}$
3 $\mathrm{K}$
4 $\mathrm{Na}$
Explanation:
The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes i.e. $\mathrm{n}=3$ and $\mathrm{l}=0$ ( for radial node $=2$ ) Hence, the valence electron of the metal must present in $3 \mathrm{~s}-\mathrm{orbital}$. The electronic configuration of the given metal is - $\begin{aligned} & { }_{29} \mathrm{Cu}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^1 \\ & { }_3 \mathrm{Li}=[\mathrm{He}] 2 \mathrm{~s}^1 \\ & { }_{19}^{\mathrm{K}}=[\mathrm{Ar}] 4 \mathrm{~s}^1 \\ & { }_{11} \mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1 \end{aligned}$ Thus, the given wave function corresponds to the valence electrons of the sodium metal.
AP- EAMCET(Medical) -2010
Structure of Atom
238854
Assertion (A) : The probability of finding an electron in a small volume around a point $(x, y$, $z$ ) at a distance ' $r$ ' from the nucleus is proportional to $\psi^2$. Reason (R) : Subatomic particles have both wave and particle nature. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $(\mathrm{A})$
2 Both (A) and (R) are true but (R) is not the correct explanation of (A)
3 Only (A) is true but (R) is not true
4 (A) is not true but (R) is true
Explanation:
The square of the wave function, $\psi^2$ is called probability density and is always positive. The probability of finding an electron at a distance ' $r$ ' from the nucleus is expressed by following relation- Probability $=4 \pi \mathrm{r}^2 \mathrm{dr} \cdot \psi^2$ $\therefore \quad$ Probability $\propto \psi^2$ The de-Broglie explains any particle shows the both nature i.e. particle and wave nature. Both (A) and (R) are true but (R) is not the correct explanation of (A).
AP - EAMCET(MEDICAL) - 2009
Structure of Atom
238855
Wave number of spectral line for a given transition is $\mathrm{x} \mathrm{cm}^{-1}$ for $\mathrm{He}^{+}$, then its value for $\mathrm{Be}^{3+}$ (isoelectronic of $\mathrm{He}^{+}$) for same transition is-
1 $\frac{x}{4} \mathrm{~cm}^{-1}$
2 $\mathrm{x} \mathrm{cm}^{-1}$
3 $4 \mathrm{x} \mathrm{cm}-1$
4 $16 \mathrm{x} \mathrm{cm}^{-1}$
Explanation:
The spectral lines of the transitions are directly proportional to square of atomic number i.e. $\mathrm{z}^2$. $\bar{v} \propto z^2$ Given, $\quad \bar{v}_1=\mathrm{x}$ Now, $\quad \frac{\bar{v}_1}{\bar{v}_2}=\frac{\mathrm{z}_1^2}{\mathrm{z}_2^2}$ For, $\mathrm{He}^{+}, \mathrm{z}_1=2$ For, $\quad \mathrm{Be}^{3+}, \mathrm{z}_2=4$ $\begin{aligned} & \frac{\mathrm{x}}{\bar{v}_2}=\frac{(4)^2}{(2)^2} \\ & \bar{v}_2=4 \mathrm{x} \mathrm{cm}^{-1} \end{aligned}$
BCECE-2013
Structure of Atom
238857
The graph between $|\psi|^2$ and $r$ (radial distance) is shown below. This represents.
1 1 s-orbital
2 2 p-orbital
3 $3 \mathrm{~s}$-orbital
4 $2 \mathrm{~s}$-orbital
Explanation:
The given probability density curve is for $2 \mathrm{~s}$ orbital because it has only one radial node. Among other given orbitals, $1 \mathrm{~s}$ and $2 \mathrm{p}$ do not have any radial node and $3 \mathrm{~s}$ has two radial nodes.
238852
The radial probability distribution curve obtained for an orbital wave function $(\psi)$ has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function $(\psi)$ correspond to ?
1 $\mathrm{Cu}$
2 $\mathrm{Li}$
3 $\mathrm{K}$
4 $\mathrm{Na}$
Explanation:
The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes i.e. $\mathrm{n}=3$ and $\mathrm{l}=0$ ( for radial node $=2$ ) Hence, the valence electron of the metal must present in $3 \mathrm{~s}-\mathrm{orbital}$. The electronic configuration of the given metal is - $\begin{aligned} & { }_{29} \mathrm{Cu}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^1 \\ & { }_3 \mathrm{Li}=[\mathrm{He}] 2 \mathrm{~s}^1 \\ & { }_{19}^{\mathrm{K}}=[\mathrm{Ar}] 4 \mathrm{~s}^1 \\ & { }_{11} \mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1 \end{aligned}$ Thus, the given wave function corresponds to the valence electrons of the sodium metal.
AP- EAMCET(Medical) -2010
Structure of Atom
238854
Assertion (A) : The probability of finding an electron in a small volume around a point $(x, y$, $z$ ) at a distance ' $r$ ' from the nucleus is proportional to $\psi^2$. Reason (R) : Subatomic particles have both wave and particle nature. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $(\mathrm{A})$
2 Both (A) and (R) are true but (R) is not the correct explanation of (A)
3 Only (A) is true but (R) is not true
4 (A) is not true but (R) is true
Explanation:
The square of the wave function, $\psi^2$ is called probability density and is always positive. The probability of finding an electron at a distance ' $r$ ' from the nucleus is expressed by following relation- Probability $=4 \pi \mathrm{r}^2 \mathrm{dr} \cdot \psi^2$ $\therefore \quad$ Probability $\propto \psi^2$ The de-Broglie explains any particle shows the both nature i.e. particle and wave nature. Both (A) and (R) are true but (R) is not the correct explanation of (A).
AP - EAMCET(MEDICAL) - 2009
Structure of Atom
238855
Wave number of spectral line for a given transition is $\mathrm{x} \mathrm{cm}^{-1}$ for $\mathrm{He}^{+}$, then its value for $\mathrm{Be}^{3+}$ (isoelectronic of $\mathrm{He}^{+}$) for same transition is-
1 $\frac{x}{4} \mathrm{~cm}^{-1}$
2 $\mathrm{x} \mathrm{cm}^{-1}$
3 $4 \mathrm{x} \mathrm{cm}-1$
4 $16 \mathrm{x} \mathrm{cm}^{-1}$
Explanation:
The spectral lines of the transitions are directly proportional to square of atomic number i.e. $\mathrm{z}^2$. $\bar{v} \propto z^2$ Given, $\quad \bar{v}_1=\mathrm{x}$ Now, $\quad \frac{\bar{v}_1}{\bar{v}_2}=\frac{\mathrm{z}_1^2}{\mathrm{z}_2^2}$ For, $\mathrm{He}^{+}, \mathrm{z}_1=2$ For, $\quad \mathrm{Be}^{3+}, \mathrm{z}_2=4$ $\begin{aligned} & \frac{\mathrm{x}}{\bar{v}_2}=\frac{(4)^2}{(2)^2} \\ & \bar{v}_2=4 \mathrm{x} \mathrm{cm}^{-1} \end{aligned}$
BCECE-2013
Structure of Atom
238857
The graph between $|\psi|^2$ and $r$ (radial distance) is shown below. This represents.
1 1 s-orbital
2 2 p-orbital
3 $3 \mathrm{~s}$-orbital
4 $2 \mathrm{~s}$-orbital
Explanation:
The given probability density curve is for $2 \mathrm{~s}$ orbital because it has only one radial node. Among other given orbitals, $1 \mathrm{~s}$ and $2 \mathrm{p}$ do not have any radial node and $3 \mathrm{~s}$ has two radial nodes.
238852
The radial probability distribution curve obtained for an orbital wave function $(\psi)$ has 3 peaks and 2 radial nodes. The valence electron of which one of the following metals does this wave function $(\psi)$ correspond to ?
1 $\mathrm{Cu}$
2 $\mathrm{Li}$
3 $\mathrm{K}$
4 $\mathrm{Na}$
Explanation:
The radial probability distribution curve obtained for an orbital wave function has 3 peaks and 2 radial nodes i.e. $\mathrm{n}=3$ and $\mathrm{l}=0$ ( for radial node $=2$ ) Hence, the valence electron of the metal must present in $3 \mathrm{~s}-\mathrm{orbital}$. The electronic configuration of the given metal is - $\begin{aligned} & { }_{29} \mathrm{Cu}=[\mathrm{Ar}] 3 \mathrm{~d}^{10}, 4 \mathrm{~s}^1 \\ & { }_3 \mathrm{Li}=[\mathrm{He}] 2 \mathrm{~s}^1 \\ & { }_{19}^{\mathrm{K}}=[\mathrm{Ar}] 4 \mathrm{~s}^1 \\ & { }_{11} \mathrm{Na}=[\mathrm{Ne}] 3 \mathrm{~s}^1 \end{aligned}$ Thus, the given wave function corresponds to the valence electrons of the sodium metal.
AP- EAMCET(Medical) -2010
Structure of Atom
238854
Assertion (A) : The probability of finding an electron in a small volume around a point $(x, y$, $z$ ) at a distance ' $r$ ' from the nucleus is proportional to $\psi^2$. Reason (R) : Subatomic particles have both wave and particle nature. The correct answer is
1 Both (A) and (R) are true and (R) is the correct explanation of $(\mathrm{A})$
2 Both (A) and (R) are true but (R) is not the correct explanation of (A)
3 Only (A) is true but (R) is not true
4 (A) is not true but (R) is true
Explanation:
The square of the wave function, $\psi^2$ is called probability density and is always positive. The probability of finding an electron at a distance ' $r$ ' from the nucleus is expressed by following relation- Probability $=4 \pi \mathrm{r}^2 \mathrm{dr} \cdot \psi^2$ $\therefore \quad$ Probability $\propto \psi^2$ The de-Broglie explains any particle shows the both nature i.e. particle and wave nature. Both (A) and (R) are true but (R) is not the correct explanation of (A).
AP - EAMCET(MEDICAL) - 2009
Structure of Atom
238855
Wave number of spectral line for a given transition is $\mathrm{x} \mathrm{cm}^{-1}$ for $\mathrm{He}^{+}$, then its value for $\mathrm{Be}^{3+}$ (isoelectronic of $\mathrm{He}^{+}$) for same transition is-
1 $\frac{x}{4} \mathrm{~cm}^{-1}$
2 $\mathrm{x} \mathrm{cm}^{-1}$
3 $4 \mathrm{x} \mathrm{cm}-1$
4 $16 \mathrm{x} \mathrm{cm}^{-1}$
Explanation:
The spectral lines of the transitions are directly proportional to square of atomic number i.e. $\mathrm{z}^2$. $\bar{v} \propto z^2$ Given, $\quad \bar{v}_1=\mathrm{x}$ Now, $\quad \frac{\bar{v}_1}{\bar{v}_2}=\frac{\mathrm{z}_1^2}{\mathrm{z}_2^2}$ For, $\mathrm{He}^{+}, \mathrm{z}_1=2$ For, $\quad \mathrm{Be}^{3+}, \mathrm{z}_2=4$ $\begin{aligned} & \frac{\mathrm{x}}{\bar{v}_2}=\frac{(4)^2}{(2)^2} \\ & \bar{v}_2=4 \mathrm{x} \mathrm{cm}^{-1} \end{aligned}$
BCECE-2013
Structure of Atom
238857
The graph between $|\psi|^2$ and $r$ (radial distance) is shown below. This represents.
1 1 s-orbital
2 2 p-orbital
3 $3 \mathrm{~s}$-orbital
4 $2 \mathrm{~s}$-orbital
Explanation:
The given probability density curve is for $2 \mathrm{~s}$ orbital because it has only one radial node. Among other given orbitals, $1 \mathrm{~s}$ and $2 \mathrm{p}$ do not have any radial node and $3 \mathrm{~s}$ has two radial nodes.
