145554
Let the series limit for Balmer series be ' $\lambda_{1}$ ', and the longest wavelength for Brackett series be ' $\lambda_{2}$ '. Then $\lambda_{1}$ and $\lambda_{2}$ are related as
1 $\lambda_{2}=1.11 \lambda_{1}$
2 $\lambda_{1}=0.09 \lambda_{2}$
3 $\lambda_{2}=0.09 \lambda_{1}$
4 $\lambda_{1}=1.11 \lambda_{2}$
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \text { for Balmer series }$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ for Bracket series So $\frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ And longest wavelength of Bracket $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{16}-\frac{1}{25}\right)=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}$ $\lambda_{2}=\frac{9}{100} \lambda_{1}=0.09 \lambda_{1}$ $\lambda_{2}=0.09 \lambda_{1}$
MHT-CET 2020
ATOMS
145555
The ratio of areas of electron orbits for the second excited state to the first excited state in hydrogen atom, is
1 $\frac{16}{81}$
2 $\frac{81}{16}$
3 $\frac{9}{4}$
4 $\frac{4}{9}$
Explanation:
B We know that, square radius of the orbit is directly proportional to the power of four of the number of state \(\mathrm{r} \propto \mathrm{n}^2\) \(\text { So, } \mathrm{r}^2 \propto \mathrm{n}^4\) Where, $\mathrm{n}$ is the number of state and area of hydrogen for second excited state $\mathrm{A}_{2} \propto \mathrm{n}^{4}$ $\mathrm{n}_{2}=3$ And area of first excited state $\mathrm{A}_{1} \propto \mathrm{n}_{1}^{4}$ $\mathrm{n}_{1}=2$ Then, the ratio of area between the electron orbits for second excited state to the first excited state. $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)^{4}=\left(\frac{3}{2}\right)^{4}=\left(\frac{81}{16}\right)$ $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{81}{16}$
MHT-CET 2020
ATOMS
145556
When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes
1 $\left(\frac{2}{3}\right)^{\text {rd }}$
2 half
3 $\left(\frac{1}{4}\right)^{\text {th }}$
4 $\left(\frac{1}{3}\right)^{\text {rd }}$
Explanation:
C We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ And kinetic energy $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}_{\mathrm{k}}}\frac{\frac{\mathrm{h}}{\mathrm{p}_{1}}}{\mathrm{~T}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\sqrt{2 \mathrm{mE}_{2}}}{\sqrt{2 \mathrm{mE}_{1}}}$ We know that, $\quad E=\frac{-13.6 Z^{2}}{n^{2}}$ Therefore, the ratio of wavelength of ground state $(n=$ 1) to $3^{\text {rd }}$ excited state $(n=4)$, $\lambda_{1} \propto \frac{1}{\mathrm{n}} \text { then } \frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4}$
MHT-CET 2020
ATOMS
145557
Let $v_{1}$ and $v_{3}$ be the frequency for series limit Balmer and Paschen series respectively. If the frequency of first line of Balmer series is $v_{2}$ then, relation between $v_{1}$ and $v_{2}$ and $v_{3}$ is
1 $v_{1}+v_{3}=v_{2}$
2 $v_{1}-v_{2}=v_{3}$
3 $v_{1}+v_{2}=v_{3}$
4 $v_{2}-v_{1}=v_{3}$
Explanation:
B We know that the wavelength of Balmer series limit $\frac{1}{\lambda_{1}}=\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right) \mathrm{R}$ Then $\quad \frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ Then using $\mathrm{c}=\mathrm{v} \lambda$ $\because \quad \frac{v_{1}}{\mathrm{c}}=\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\mathrm{v}_{1}=\frac{\mathrm{Rc}}{4}$ Similarly, wavelength of Paschan series limit $\frac{1}{\lambda_{3}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{3}}=\frac{\mathrm{R}}{9}=\frac{\mathrm{v}_{3}}{\mathrm{c}}$ $v_{3}=\frac{\mathrm{Rc}}{9}$ Similarly, wavelength of first of Balmer series $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{9}$ $v_{2}=\frac{\mathrm{Rc}}{4}-\frac{\mathrm{Rc}}{9}=v_{1}-v_{3}$ $v_{3}=v_{1}-v_{2}$
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ATOMS
145554
Let the series limit for Balmer series be ' $\lambda_{1}$ ', and the longest wavelength for Brackett series be ' $\lambda_{2}$ '. Then $\lambda_{1}$ and $\lambda_{2}$ are related as
1 $\lambda_{2}=1.11 \lambda_{1}$
2 $\lambda_{1}=0.09 \lambda_{2}$
3 $\lambda_{2}=0.09 \lambda_{1}$
4 $\lambda_{1}=1.11 \lambda_{2}$
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \text { for Balmer series }$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ for Bracket series So $\frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ And longest wavelength of Bracket $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{16}-\frac{1}{25}\right)=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}$ $\lambda_{2}=\frac{9}{100} \lambda_{1}=0.09 \lambda_{1}$ $\lambda_{2}=0.09 \lambda_{1}$
MHT-CET 2020
ATOMS
145555
The ratio of areas of electron orbits for the second excited state to the first excited state in hydrogen atom, is
1 $\frac{16}{81}$
2 $\frac{81}{16}$
3 $\frac{9}{4}$
4 $\frac{4}{9}$
Explanation:
B We know that, square radius of the orbit is directly proportional to the power of four of the number of state \(\mathrm{r} \propto \mathrm{n}^2\) \(\text { So, } \mathrm{r}^2 \propto \mathrm{n}^4\) Where, $\mathrm{n}$ is the number of state and area of hydrogen for second excited state $\mathrm{A}_{2} \propto \mathrm{n}^{4}$ $\mathrm{n}_{2}=3$ And area of first excited state $\mathrm{A}_{1} \propto \mathrm{n}_{1}^{4}$ $\mathrm{n}_{1}=2$ Then, the ratio of area between the electron orbits for second excited state to the first excited state. $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)^{4}=\left(\frac{3}{2}\right)^{4}=\left(\frac{81}{16}\right)$ $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{81}{16}$
MHT-CET 2020
ATOMS
145556
When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes
1 $\left(\frac{2}{3}\right)^{\text {rd }}$
2 half
3 $\left(\frac{1}{4}\right)^{\text {th }}$
4 $\left(\frac{1}{3}\right)^{\text {rd }}$
Explanation:
C We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ And kinetic energy $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}_{\mathrm{k}}}\frac{\frac{\mathrm{h}}{\mathrm{p}_{1}}}{\mathrm{~T}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\sqrt{2 \mathrm{mE}_{2}}}{\sqrt{2 \mathrm{mE}_{1}}}$ We know that, $\quad E=\frac{-13.6 Z^{2}}{n^{2}}$ Therefore, the ratio of wavelength of ground state $(n=$ 1) to $3^{\text {rd }}$ excited state $(n=4)$, $\lambda_{1} \propto \frac{1}{\mathrm{n}} \text { then } \frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4}$
MHT-CET 2020
ATOMS
145557
Let $v_{1}$ and $v_{3}$ be the frequency for series limit Balmer and Paschen series respectively. If the frequency of first line of Balmer series is $v_{2}$ then, relation between $v_{1}$ and $v_{2}$ and $v_{3}$ is
1 $v_{1}+v_{3}=v_{2}$
2 $v_{1}-v_{2}=v_{3}$
3 $v_{1}+v_{2}=v_{3}$
4 $v_{2}-v_{1}=v_{3}$
Explanation:
B We know that the wavelength of Balmer series limit $\frac{1}{\lambda_{1}}=\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right) \mathrm{R}$ Then $\quad \frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ Then using $\mathrm{c}=\mathrm{v} \lambda$ $\because \quad \frac{v_{1}}{\mathrm{c}}=\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\mathrm{v}_{1}=\frac{\mathrm{Rc}}{4}$ Similarly, wavelength of Paschan series limit $\frac{1}{\lambda_{3}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{3}}=\frac{\mathrm{R}}{9}=\frac{\mathrm{v}_{3}}{\mathrm{c}}$ $v_{3}=\frac{\mathrm{Rc}}{9}$ Similarly, wavelength of first of Balmer series $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{9}$ $v_{2}=\frac{\mathrm{Rc}}{4}-\frac{\mathrm{Rc}}{9}=v_{1}-v_{3}$ $v_{3}=v_{1}-v_{2}$
145554
Let the series limit for Balmer series be ' $\lambda_{1}$ ', and the longest wavelength for Brackett series be ' $\lambda_{2}$ '. Then $\lambda_{1}$ and $\lambda_{2}$ are related as
1 $\lambda_{2}=1.11 \lambda_{1}$
2 $\lambda_{1}=0.09 \lambda_{2}$
3 $\lambda_{2}=0.09 \lambda_{1}$
4 $\lambda_{1}=1.11 \lambda_{2}$
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \text { for Balmer series }$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ for Bracket series So $\frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ And longest wavelength of Bracket $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{16}-\frac{1}{25}\right)=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}$ $\lambda_{2}=\frac{9}{100} \lambda_{1}=0.09 \lambda_{1}$ $\lambda_{2}=0.09 \lambda_{1}$
MHT-CET 2020
ATOMS
145555
The ratio of areas of electron orbits for the second excited state to the first excited state in hydrogen atom, is
1 $\frac{16}{81}$
2 $\frac{81}{16}$
3 $\frac{9}{4}$
4 $\frac{4}{9}$
Explanation:
B We know that, square radius of the orbit is directly proportional to the power of four of the number of state \(\mathrm{r} \propto \mathrm{n}^2\) \(\text { So, } \mathrm{r}^2 \propto \mathrm{n}^4\) Where, $\mathrm{n}$ is the number of state and area of hydrogen for second excited state $\mathrm{A}_{2} \propto \mathrm{n}^{4}$ $\mathrm{n}_{2}=3$ And area of first excited state $\mathrm{A}_{1} \propto \mathrm{n}_{1}^{4}$ $\mathrm{n}_{1}=2$ Then, the ratio of area between the electron orbits for second excited state to the first excited state. $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)^{4}=\left(\frac{3}{2}\right)^{4}=\left(\frac{81}{16}\right)$ $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{81}{16}$
MHT-CET 2020
ATOMS
145556
When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes
1 $\left(\frac{2}{3}\right)^{\text {rd }}$
2 half
3 $\left(\frac{1}{4}\right)^{\text {th }}$
4 $\left(\frac{1}{3}\right)^{\text {rd }}$
Explanation:
C We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ And kinetic energy $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}_{\mathrm{k}}}\frac{\frac{\mathrm{h}}{\mathrm{p}_{1}}}{\mathrm{~T}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\sqrt{2 \mathrm{mE}_{2}}}{\sqrt{2 \mathrm{mE}_{1}}}$ We know that, $\quad E=\frac{-13.6 Z^{2}}{n^{2}}$ Therefore, the ratio of wavelength of ground state $(n=$ 1) to $3^{\text {rd }}$ excited state $(n=4)$, $\lambda_{1} \propto \frac{1}{\mathrm{n}} \text { then } \frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4}$
MHT-CET 2020
ATOMS
145557
Let $v_{1}$ and $v_{3}$ be the frequency for series limit Balmer and Paschen series respectively. If the frequency of first line of Balmer series is $v_{2}$ then, relation between $v_{1}$ and $v_{2}$ and $v_{3}$ is
1 $v_{1}+v_{3}=v_{2}$
2 $v_{1}-v_{2}=v_{3}$
3 $v_{1}+v_{2}=v_{3}$
4 $v_{2}-v_{1}=v_{3}$
Explanation:
B We know that the wavelength of Balmer series limit $\frac{1}{\lambda_{1}}=\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right) \mathrm{R}$ Then $\quad \frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ Then using $\mathrm{c}=\mathrm{v} \lambda$ $\because \quad \frac{v_{1}}{\mathrm{c}}=\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\mathrm{v}_{1}=\frac{\mathrm{Rc}}{4}$ Similarly, wavelength of Paschan series limit $\frac{1}{\lambda_{3}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{3}}=\frac{\mathrm{R}}{9}=\frac{\mathrm{v}_{3}}{\mathrm{c}}$ $v_{3}=\frac{\mathrm{Rc}}{9}$ Similarly, wavelength of first of Balmer series $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{9}$ $v_{2}=\frac{\mathrm{Rc}}{4}-\frac{\mathrm{Rc}}{9}=v_{1}-v_{3}$ $v_{3}=v_{1}-v_{2}$
145554
Let the series limit for Balmer series be ' $\lambda_{1}$ ', and the longest wavelength for Brackett series be ' $\lambda_{2}$ '. Then $\lambda_{1}$ and $\lambda_{2}$ are related as
1 $\lambda_{2}=1.11 \lambda_{1}$
2 $\lambda_{1}=0.09 \lambda_{2}$
3 $\lambda_{2}=0.09 \lambda_{1}$
4 $\lambda_{1}=1.11 \lambda_{2}$
Explanation:
C We know that, $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{\mathrm{n}^{2}}\right) \text { for Balmer series }$ $\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{4^{2}}-\frac{1}{\mathrm{n}^{2}}\right)$ for Bracket series So $\frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ And longest wavelength of Bracket $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{16}-\frac{1}{25}\right)=\mathrm{R}\left[\frac{25-16}{16 \times 25}\right]=\frac{9 \mathrm{R}}{16 \times 25}$ $\frac{\lambda_{1}}{\lambda_{2}}=\frac{\mathrm{R}}{4} \times \frac{16 \times 25}{9 \mathrm{R}}=\frac{100}{9}$ $\lambda_{2}=\frac{9}{100} \lambda_{1}=0.09 \lambda_{1}$ $\lambda_{2}=0.09 \lambda_{1}$
MHT-CET 2020
ATOMS
145555
The ratio of areas of electron orbits for the second excited state to the first excited state in hydrogen atom, is
1 $\frac{16}{81}$
2 $\frac{81}{16}$
3 $\frac{9}{4}$
4 $\frac{4}{9}$
Explanation:
B We know that, square radius of the orbit is directly proportional to the power of four of the number of state \(\mathrm{r} \propto \mathrm{n}^2\) \(\text { So, } \mathrm{r}^2 \propto \mathrm{n}^4\) Where, $\mathrm{n}$ is the number of state and area of hydrogen for second excited state $\mathrm{A}_{2} \propto \mathrm{n}^{4}$ $\mathrm{n}_{2}=3$ And area of first excited state $\mathrm{A}_{1} \propto \mathrm{n}_{1}^{4}$ $\mathrm{n}_{1}=2$ Then, the ratio of area between the electron orbits for second excited state to the first excited state. $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\left(\frac{\mathrm{n}_{2}}{\mathrm{n}_{1}}\right)^{4}=\left(\frac{3}{2}\right)^{4}=\left(\frac{81}{16}\right)$ $\frac{\mathrm{A}_{2}}{\mathrm{~A}_{1}}=\frac{81}{16}$
MHT-CET 2020
ATOMS
145556
When an electron in hydrogen atom jumps from third excited state to the ground state, the de-Broglie wavelength associated with the electron becomes
1 $\left(\frac{2}{3}\right)^{\text {rd }}$
2 half
3 $\left(\frac{1}{4}\right)^{\text {th }}$
4 $\left(\frac{1}{3}\right)^{\text {rd }}$
Explanation:
C We know that, de-Broglie wavelength $(\lambda)=\frac{h}{p}$ And kinetic energy $\mathrm{E}_{\mathrm{K}}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{p}=\sqrt{2 \mathrm{mE}_{\mathrm{k}}}\frac{\frac{\mathrm{h}}{\mathrm{p}_{1}}}{\mathrm{~T}}=\frac{\mathrm{p}_{2}}{\mathrm{p}_{1}}=\frac{\sqrt{2 \mathrm{mE}_{2}}}{\sqrt{2 \mathrm{mE}_{1}}}$ We know that, $\quad E=\frac{-13.6 Z^{2}}{n^{2}}$ Therefore, the ratio of wavelength of ground state $(n=$ 1) to $3^{\text {rd }}$ excited state $(n=4)$, $\lambda_{1} \propto \frac{1}{\mathrm{n}} \text { then } \frac{\lambda_{1}}{\lambda_{4}}=\frac{1}{4}$
MHT-CET 2020
ATOMS
145557
Let $v_{1}$ and $v_{3}$ be the frequency for series limit Balmer and Paschen series respectively. If the frequency of first line of Balmer series is $v_{2}$ then, relation between $v_{1}$ and $v_{2}$ and $v_{3}$ is
1 $v_{1}+v_{3}=v_{2}$
2 $v_{1}-v_{2}=v_{3}$
3 $v_{1}+v_{2}=v_{3}$
4 $v_{2}-v_{1}=v_{3}$
Explanation:
B We know that the wavelength of Balmer series limit $\frac{1}{\lambda_{1}}=\left(\frac{1}{2^{2}}-\frac{1}{\infty^{2}}\right) \mathrm{R}$ Then $\quad \frac{1}{\lambda_{1}}=\frac{\mathrm{R}}{4}$ Then using $\mathrm{c}=\mathrm{v} \lambda$ $\because \quad \frac{v_{1}}{\mathrm{c}}=\frac{1}{\lambda}=\frac{\mathrm{R}}{4}$ $\mathrm{v}_{1}=\frac{\mathrm{Rc}}{4}$ Similarly, wavelength of Paschan series limit $\frac{1}{\lambda_{3}}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{\infty^{2}}\right)$ $\frac{1}{\lambda_{3}}=\frac{\mathrm{R}}{9}=\frac{\mathrm{v}_{3}}{\mathrm{c}}$ $v_{3}=\frac{\mathrm{Rc}}{9}$ Similarly, wavelength of first of Balmer series $\frac{1}{\lambda_{2}}=\mathrm{R}\left(\frac{1}{2^{2}}-\frac{1}{3^{2}}\right)=\frac{\mathrm{R}}{4}-\frac{\mathrm{R}}{9}$ $v_{2}=\frac{\mathrm{Rc}}{4}-\frac{\mathrm{Rc}}{9}=v_{1}-v_{3}$ $v_{3}=v_{1}-v_{2}$
