145251
The energy levels of a hypothetical one electron atom system are given by $E_{n}=-\frac{16}{n^{2}} e V \text {, where } n=1,2,3, \ldots .$ The wavelength of emitted photon corresponding to transition from first excited level to ground level is about
1 $1035 \AA$
2 $1220 \AA$
3 $3650 \AA$
4 $690 \AA$
Explanation:
A Given, $\mathrm{E}_{\mathrm{n}}=\frac{-16}{\mathrm{n}^{2}} \mathrm{eV}$ Where, $\mathrm{n}=1,2,3, \ldots \ldots$ First excited level means, $\mathrm{n}=2$ and ground level, $\mathrm{n}=1$ Then, $\mathrm{E}_{1}=\frac{-16}{1} \mathrm{eV}=-16 \mathrm{eV}$ $\mathrm{E}_{2}=\frac{-16}{2^{2}} \mathrm{eV}=-4 \mathrm{eV}$ $\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}=-4-(-16)=12 \mathrm{eV}$ We know that, $\lambda=\frac{12420}{\Delta \mathrm{E}}=\frac{12420}{12}=1035 \AA$ $\lambda=1035 \AA$
UPSEE - 2017
ATOMS
145256
What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus?
1 $1: 3$
2 $1: 2$
3 $1: 8$
4 $2: 6$
Explanation:
B Given, Mass of $\operatorname{Li}(\mathrm{A})=7$ Mass of $\mathrm{Fe}(\mathrm{A})=56$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $R_{0}=$ constant and value of $R_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Radii of $\mathrm{Li}=\mathrm{R}_{0}(7)^{1 / 3}$ Radii of $\mathrm{Fe}=\mathrm{R}_{0}(56)^{1 / 3}$ $\because \quad \frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\frac{\mathrm{R}_{0}(7)^{1 / 3}}{\mathrm{R}_{0}(56)^{1 / 3}}=\left(\frac{7}{56}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\left(\frac{1}{8}\right)^{3}=\frac{1}{2}$ $\mathrm{R}_{\mathrm{Li}}: \mathrm{R}_{\mathrm{Fe}}=1: 2$
CG PET- 2008
ATOMS
145257
An electron is at ground state of the $H$ atom. Minimum energy required to excite the $\mathrm{H}$ atom into second excited state is
1 $13.6 \mathrm{eV}$
2 $12.1 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $3.4 \mathrm{eV}$
Explanation:
B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level, $E=\frac{-13.6 Z^{2}}{n^{2}}$ Where, $\mathrm{Z}$ is atomic number and $\mathrm{n}$ is energy state. For $\mathrm{Z}=1, \mathrm{n}=1$ $\mathrm{E}_{1}=\frac{-13.6 \times 1^{2}}{1^{2}}=-13.6 \mathrm{eV}$ For an electron jump from first excited state to second. $\therefore \mathrm{Z}=1, \mathrm{n}=3$ $\mathrm{E}_{3}=\frac{-13.6 \times 1^{2}}{3^{2}}=-1.5 \mathrm{eV}$ Therefore, $\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}$ $\Delta \mathrm{E}=-1.5-(-13.6)$ $\Delta \mathrm{E}=12.1 \mathrm{eV}$
UPSEE - 2016
ATOMS
145261
The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_{4}^{9} \mathrm{Be}$. The number of nucleons in Ge are :
1 73
2 74
3 75
4 72
Explanation:
D Given, Radius of germanium $=2$ times radius of Beryllium $\mathrm{R}_{\mathrm{Ge}}=2 \mathrm{R}_{\mathrm{Be}}$ Mass of Beryllium (A) $=9$ Mass of germanium $(\mathrm{A})=$ ? We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant and value of $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{\mathrm{R}_{\mathrm{Ge}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ From equation (i), we get- $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{2 \mathrm{R}_{\mathrm{Be}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=2 \times(9)^{1 / 3}$ $\mathrm{A}=2^{3} \times 9$ $\therefore$ The number of nucleons in $\mathrm{Ge}$ is 72 .
145251
The energy levels of a hypothetical one electron atom system are given by $E_{n}=-\frac{16}{n^{2}} e V \text {, where } n=1,2,3, \ldots .$ The wavelength of emitted photon corresponding to transition from first excited level to ground level is about
1 $1035 \AA$
2 $1220 \AA$
3 $3650 \AA$
4 $690 \AA$
Explanation:
A Given, $\mathrm{E}_{\mathrm{n}}=\frac{-16}{\mathrm{n}^{2}} \mathrm{eV}$ Where, $\mathrm{n}=1,2,3, \ldots \ldots$ First excited level means, $\mathrm{n}=2$ and ground level, $\mathrm{n}=1$ Then, $\mathrm{E}_{1}=\frac{-16}{1} \mathrm{eV}=-16 \mathrm{eV}$ $\mathrm{E}_{2}=\frac{-16}{2^{2}} \mathrm{eV}=-4 \mathrm{eV}$ $\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}=-4-(-16)=12 \mathrm{eV}$ We know that, $\lambda=\frac{12420}{\Delta \mathrm{E}}=\frac{12420}{12}=1035 \AA$ $\lambda=1035 \AA$
UPSEE - 2017
ATOMS
145256
What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus?
1 $1: 3$
2 $1: 2$
3 $1: 8$
4 $2: 6$
Explanation:
B Given, Mass of $\operatorname{Li}(\mathrm{A})=7$ Mass of $\mathrm{Fe}(\mathrm{A})=56$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $R_{0}=$ constant and value of $R_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Radii of $\mathrm{Li}=\mathrm{R}_{0}(7)^{1 / 3}$ Radii of $\mathrm{Fe}=\mathrm{R}_{0}(56)^{1 / 3}$ $\because \quad \frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\frac{\mathrm{R}_{0}(7)^{1 / 3}}{\mathrm{R}_{0}(56)^{1 / 3}}=\left(\frac{7}{56}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\left(\frac{1}{8}\right)^{3}=\frac{1}{2}$ $\mathrm{R}_{\mathrm{Li}}: \mathrm{R}_{\mathrm{Fe}}=1: 2$
CG PET- 2008
ATOMS
145257
An electron is at ground state of the $H$ atom. Minimum energy required to excite the $\mathrm{H}$ atom into second excited state is
1 $13.6 \mathrm{eV}$
2 $12.1 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $3.4 \mathrm{eV}$
Explanation:
B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level, $E=\frac{-13.6 Z^{2}}{n^{2}}$ Where, $\mathrm{Z}$ is atomic number and $\mathrm{n}$ is energy state. For $\mathrm{Z}=1, \mathrm{n}=1$ $\mathrm{E}_{1}=\frac{-13.6 \times 1^{2}}{1^{2}}=-13.6 \mathrm{eV}$ For an electron jump from first excited state to second. $\therefore \mathrm{Z}=1, \mathrm{n}=3$ $\mathrm{E}_{3}=\frac{-13.6 \times 1^{2}}{3^{2}}=-1.5 \mathrm{eV}$ Therefore, $\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}$ $\Delta \mathrm{E}=-1.5-(-13.6)$ $\Delta \mathrm{E}=12.1 \mathrm{eV}$
UPSEE - 2016
ATOMS
145261
The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_{4}^{9} \mathrm{Be}$. The number of nucleons in Ge are :
1 73
2 74
3 75
4 72
Explanation:
D Given, Radius of germanium $=2$ times radius of Beryllium $\mathrm{R}_{\mathrm{Ge}}=2 \mathrm{R}_{\mathrm{Be}}$ Mass of Beryllium (A) $=9$ Mass of germanium $(\mathrm{A})=$ ? We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant and value of $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{\mathrm{R}_{\mathrm{Ge}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ From equation (i), we get- $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{2 \mathrm{R}_{\mathrm{Be}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=2 \times(9)^{1 / 3}$ $\mathrm{A}=2^{3} \times 9$ $\therefore$ The number of nucleons in $\mathrm{Ge}$ is 72 .
145251
The energy levels of a hypothetical one electron atom system are given by $E_{n}=-\frac{16}{n^{2}} e V \text {, where } n=1,2,3, \ldots .$ The wavelength of emitted photon corresponding to transition from first excited level to ground level is about
1 $1035 \AA$
2 $1220 \AA$
3 $3650 \AA$
4 $690 \AA$
Explanation:
A Given, $\mathrm{E}_{\mathrm{n}}=\frac{-16}{\mathrm{n}^{2}} \mathrm{eV}$ Where, $\mathrm{n}=1,2,3, \ldots \ldots$ First excited level means, $\mathrm{n}=2$ and ground level, $\mathrm{n}=1$ Then, $\mathrm{E}_{1}=\frac{-16}{1} \mathrm{eV}=-16 \mathrm{eV}$ $\mathrm{E}_{2}=\frac{-16}{2^{2}} \mathrm{eV}=-4 \mathrm{eV}$ $\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}=-4-(-16)=12 \mathrm{eV}$ We know that, $\lambda=\frac{12420}{\Delta \mathrm{E}}=\frac{12420}{12}=1035 \AA$ $\lambda=1035 \AA$
UPSEE - 2017
ATOMS
145256
What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus?
1 $1: 3$
2 $1: 2$
3 $1: 8$
4 $2: 6$
Explanation:
B Given, Mass of $\operatorname{Li}(\mathrm{A})=7$ Mass of $\mathrm{Fe}(\mathrm{A})=56$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $R_{0}=$ constant and value of $R_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Radii of $\mathrm{Li}=\mathrm{R}_{0}(7)^{1 / 3}$ Radii of $\mathrm{Fe}=\mathrm{R}_{0}(56)^{1 / 3}$ $\because \quad \frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\frac{\mathrm{R}_{0}(7)^{1 / 3}}{\mathrm{R}_{0}(56)^{1 / 3}}=\left(\frac{7}{56}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\left(\frac{1}{8}\right)^{3}=\frac{1}{2}$ $\mathrm{R}_{\mathrm{Li}}: \mathrm{R}_{\mathrm{Fe}}=1: 2$
CG PET- 2008
ATOMS
145257
An electron is at ground state of the $H$ atom. Minimum energy required to excite the $\mathrm{H}$ atom into second excited state is
1 $13.6 \mathrm{eV}$
2 $12.1 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $3.4 \mathrm{eV}$
Explanation:
B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level, $E=\frac{-13.6 Z^{2}}{n^{2}}$ Where, $\mathrm{Z}$ is atomic number and $\mathrm{n}$ is energy state. For $\mathrm{Z}=1, \mathrm{n}=1$ $\mathrm{E}_{1}=\frac{-13.6 \times 1^{2}}{1^{2}}=-13.6 \mathrm{eV}$ For an electron jump from first excited state to second. $\therefore \mathrm{Z}=1, \mathrm{n}=3$ $\mathrm{E}_{3}=\frac{-13.6 \times 1^{2}}{3^{2}}=-1.5 \mathrm{eV}$ Therefore, $\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}$ $\Delta \mathrm{E}=-1.5-(-13.6)$ $\Delta \mathrm{E}=12.1 \mathrm{eV}$
UPSEE - 2016
ATOMS
145261
The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_{4}^{9} \mathrm{Be}$. The number of nucleons in Ge are :
1 73
2 74
3 75
4 72
Explanation:
D Given, Radius of germanium $=2$ times radius of Beryllium $\mathrm{R}_{\mathrm{Ge}}=2 \mathrm{R}_{\mathrm{Be}}$ Mass of Beryllium (A) $=9$ Mass of germanium $(\mathrm{A})=$ ? We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant and value of $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{\mathrm{R}_{\mathrm{Ge}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ From equation (i), we get- $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{2 \mathrm{R}_{\mathrm{Be}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=2 \times(9)^{1 / 3}$ $\mathrm{A}=2^{3} \times 9$ $\therefore$ The number of nucleons in $\mathrm{Ge}$ is 72 .
145251
The energy levels of a hypothetical one electron atom system are given by $E_{n}=-\frac{16}{n^{2}} e V \text {, where } n=1,2,3, \ldots .$ The wavelength of emitted photon corresponding to transition from first excited level to ground level is about
1 $1035 \AA$
2 $1220 \AA$
3 $3650 \AA$
4 $690 \AA$
Explanation:
A Given, $\mathrm{E}_{\mathrm{n}}=\frac{-16}{\mathrm{n}^{2}} \mathrm{eV}$ Where, $\mathrm{n}=1,2,3, \ldots \ldots$ First excited level means, $\mathrm{n}=2$ and ground level, $\mathrm{n}=1$ Then, $\mathrm{E}_{1}=\frac{-16}{1} \mathrm{eV}=-16 \mathrm{eV}$ $\mathrm{E}_{2}=\frac{-16}{2^{2}} \mathrm{eV}=-4 \mathrm{eV}$ $\Delta \mathrm{E}=\mathrm{E}_{2}-\mathrm{E}_{1}=-4-(-16)=12 \mathrm{eV}$ We know that, $\lambda=\frac{12420}{\Delta \mathrm{E}}=\frac{12420}{12}=1035 \AA$ $\lambda=1035 \AA$
UPSEE - 2017
ATOMS
145256
What will be ratio of radii of $\mathrm{Li}^{7}$ nucleus to $\mathrm{Fe}^{56}$ nucleus?
1 $1: 3$
2 $1: 2$
3 $1: 8$
4 $2: 6$
Explanation:
B Given, Mass of $\operatorname{Li}(\mathrm{A})=7$ Mass of $\mathrm{Fe}(\mathrm{A})=56$ We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $R_{0}=$ constant and value of $R_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Radii of $\mathrm{Li}=\mathrm{R}_{0}(7)^{1 / 3}$ Radii of $\mathrm{Fe}=\mathrm{R}_{0}(56)^{1 / 3}$ $\because \quad \frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\frac{\mathrm{R}_{0}(7)^{1 / 3}}{\mathrm{R}_{0}(56)^{1 / 3}}=\left(\frac{7}{56}\right)^{1 / 3}$ $\frac{\mathrm{R}_{\mathrm{Li}}}{\mathrm{R}_{\mathrm{Fe}}}=\left(\frac{1}{8}\right)^{3}=\frac{1}{2}$ $\mathrm{R}_{\mathrm{Li}}: \mathrm{R}_{\mathrm{Fe}}=1: 2$
CG PET- 2008
ATOMS
145257
An electron is at ground state of the $H$ atom. Minimum energy required to excite the $\mathrm{H}$ atom into second excited state is
1 $13.6 \mathrm{eV}$
2 $12.1 \mathrm{eV}$
3 $10.2 \mathrm{eV}$
4 $3.4 \mathrm{eV}$
Explanation:
B Hydrogen atom having only one electron, then the energy required to excite hydrogen atom will be equal to the energy required to excite one electron to second level, $E=\frac{-13.6 Z^{2}}{n^{2}}$ Where, $\mathrm{Z}$ is atomic number and $\mathrm{n}$ is energy state. For $\mathrm{Z}=1, \mathrm{n}=1$ $\mathrm{E}_{1}=\frac{-13.6 \times 1^{2}}{1^{2}}=-13.6 \mathrm{eV}$ For an electron jump from first excited state to second. $\therefore \mathrm{Z}=1, \mathrm{n}=3$ $\mathrm{E}_{3}=\frac{-13.6 \times 1^{2}}{3^{2}}=-1.5 \mathrm{eV}$ Therefore, $\Delta \mathrm{E}=\mathrm{E}_{3}-\mathrm{E}_{1}$ $\Delta \mathrm{E}=-1.5-(-13.6)$ $\Delta \mathrm{E}=12.1 \mathrm{eV}$
UPSEE - 2016
ATOMS
145261
The radius of germanium (Ge) nuclide is measured to be twice the radius of ${ }_{4}^{9} \mathrm{Be}$. The number of nucleons in Ge are :
1 73
2 74
3 75
4 72
Explanation:
D Given, Radius of germanium $=2$ times radius of Beryllium $\mathrm{R}_{\mathrm{Ge}}=2 \mathrm{R}_{\mathrm{Be}}$ Mass of Beryllium (A) $=9$ Mass of germanium $(\mathrm{A})=$ ? We know that, Radius of nucleus, $\mathrm{R}=\mathrm{R}_{0}(\mathrm{~A})^{1 / 3}$ Where, $\mathrm{R}_{0}=$ Constant and value of $\mathrm{R}_{0}=1.2 \times 10^{-15} \mathrm{~m}$ Or $\quad \mathrm{R} \propto(\mathrm{A})^{1 / 3}$ $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{\mathrm{R}_{\mathrm{Ge}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ From equation (i), we get- $\therefore \quad \frac{\mathrm{R}_{\mathrm{Be}}}{2 \mathrm{R}_{\mathrm{Be}}}=\frac{(9)^{1 / 3}}{(\mathrm{~A})^{1 / 3}}$ $(\mathrm{A})^{1 / 3}=2 \times(9)^{1 / 3}$ $\mathrm{A}=2^{3} \times 9$ $\therefore$ The number of nucleons in $\mathrm{Ge}$ is 72 .
