145312 An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ energy state. The energy released in this process, will be $($ Given $\operatorname{Rch}=13.6 \mathrm{eV})$
Where $R=$ Rydberg constant, $c=$ speed of light in vacuum, $h=$ Planck's constant.

145313 In accordance with the Bohr's model. Find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} \mathrm{~m}$ with orbital speed $3 \times 10^{4} \mathrm{~m} / \mathrm{s}$. (Mass of earth $=6 \times 10^{24} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34}$ J.s)

145315 The radius of hydrogen atom in the ground state is $0.53 \AA$. After collision with an electron, it is found to have a radius of $2.12 \AA$, the principle quantum number ' $n$ ' of the final state of the atom is

145317 In Bohr atom model, the total energy of the electron in hydrogen atom is $-3.4 \mathrm{eV}$. Then its angular momentum about the nucleus of the atom is $(h=$ Planck's constant $)$

145312 An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ energy state. The energy released in this process, will be $($ Given $\operatorname{Rch}=13.6 \mathrm{eV})$
Where $R=$ Rydberg constant, $c=$ speed of light in vacuum, $h=$ Planck's constant.

145313 In accordance with the Bohr's model. Find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} \mathrm{~m}$ with orbital speed $3 \times 10^{4} \mathrm{~m} / \mathrm{s}$. (Mass of earth $=6 \times 10^{24} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34}$ J.s)

145315 The radius of hydrogen atom in the ground state is $0.53 \AA$. After collision with an electron, it is found to have a radius of $2.12 \AA$, the principle quantum number ' $n$ ' of the final state of the atom is

145317 In Bohr atom model, the total energy of the electron in hydrogen atom is $-3.4 \mathrm{eV}$. Then its angular momentum about the nucleus of the atom is $(h=$ Planck's constant $)$

145312 An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ energy state. The energy released in this process, will be $($ Given $\operatorname{Rch}=13.6 \mathrm{eV})$
Where $R=$ Rydberg constant, $c=$ speed of light in vacuum, $h=$ Planck's constant.

145313 In accordance with the Bohr's model. Find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} \mathrm{~m}$ with orbital speed $3 \times 10^{4} \mathrm{~m} / \mathrm{s}$. (Mass of earth $=6 \times 10^{24} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34}$ J.s)

145315 The radius of hydrogen atom in the ground state is $0.53 \AA$. After collision with an electron, it is found to have a radius of $2.12 \AA$, the principle quantum number ' $n$ ' of the final state of the atom is

145317 In Bohr atom model, the total energy of the electron in hydrogen atom is $-3.4 \mathrm{eV}$. Then its angular momentum about the nucleus of the atom is $(h=$ Planck's constant $)$

145312 An electron of a hydrogen like atom, having $Z=4$, jumps from $4^{\text {th }}$ energy state to $2^{\text {nd }}$ energy state. The energy released in this process, will be $($ Given $\operatorname{Rch}=13.6 \mathrm{eV})$
Where $R=$ Rydberg constant, $c=$ speed of light in vacuum, $h=$ Planck's constant.

145313 In accordance with the Bohr's model. Find the quantum number that characterizes the earth's revolution around the sun in an orbit of radius $1.5 \times 10^{11} \mathrm{~m}$ with orbital speed $3 \times 10^{4} \mathrm{~m} / \mathrm{s}$. (Mass of earth $=6 \times 10^{24} \mathrm{~kg}, \mathrm{~h}=6.625 \times 10^{-34}$ J.s)

145315 The radius of hydrogen atom in the ground state is $0.53 \AA$. After collision with an electron, it is found to have a radius of $2.12 \AA$, the principle quantum number ' $n$ ' of the final state of the atom is

145317 In Bohr atom model, the total energy of the electron in hydrogen atom is $-3.4 \mathrm{eV}$. Then its angular momentum about the nucleus of the atom is $(h=$ Planck's constant $)$