Apply Bohr’s atomic model to a lithium atom. Assuming that its two $K$-shell electrons are too close to nucleus such that nucleus and $K$-shell electron act as a nucleus of effective positive charge equivalent to electron. The ionization energy of its outermost electron is......$eV$
Apply Bohr’s atomic model to a lithium atom. Assuming that its two $K$-shell electrons are too close to nucleus such that nucleus and $K$-shell electron act as a nucleus of effective positive charge equivalent to electron. The ionization energy of its outermost electron is......$eV$
- A
$30.6$
- B
$3.4$
- C
$32.4$
- D
$13.6$
Similar Questions
The absorption transitions between the first and the fourth energy states of hydrogen atom are $3.$ The emission transitions between these states will be
The absorption transitions between the first and the fourth energy states of hydrogen atom are $3.$ The emission transitions between these states will be
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom $\left(\sim 10^{-10} \;m \right)$
$(a)$ Construct a quantity with the dimensions of length from the fundamental constants $e, m_{e},$ and $c .$ Determine its numerical value.
$(b)$ You will find that the length obtained in $(a)$ is many orders of magnitude smaller than the atomic dimensions. Further, it involves $c .$ But energies of atoms are mostly in non-relativistic domain where $c$ is not expected to play any role. This is what may have suggested Bohr to discard $c$ and look for 'something else' to get the right atomic size. Now, the Planck's constant $h$ had already made its appearance elsewhere. Bohr's great insight lay in recognising that $h, m_{e},$ and $e$ will yield the right atomic size. Construct a quantity with the dimension of length from $h m_e$, and $e$ and confirm that its numerical value has indeed the correct order of magnitude.
Classically, an electron can be in any orbit around the nucleus of an atom. Then what determines the typical atomic size? Why is an atom not, say, thousand times bigger than its typical size? The question had greatly puzzled Bohr before he arrived at his famous model of the atom that you have learnt in the text. To simulate what he might well have done before his discovery, let us play as follows with the basic constants of nature and see if we can get a quantity with the dimensions of length that is roughly equal to the known size of an atom $\left(\sim 10^{-10} \;m \right)$
$(a)$ Construct a quantity with the dimensions of length from the fundamental constants $e, m_{e},$ and $c .$ Determine its numerical value.
$(b)$ You will find that the length obtained in $(a)$ is many orders of magnitude smaller than the atomic dimensions. Further, it involves $c .$ But energies of atoms are mostly in non-relativistic domain where $c$ is not expected to play any role. This is what may have suggested Bohr to discard $c$ and look for 'something else' to get the right atomic size. Now, the Planck's constant $h$ had already made its appearance elsewhere. Bohr's great insight lay in recognising that $h, m_{e},$ and $e$ will yield the right atomic size. Construct a quantity with the dimension of length from $h m_e$, and $e$ and confirm that its numerical value has indeed the correct order of magnitude.
Explain Rutherford's argument for scattered $\alpha $ -particles.
Explain Rutherford's argument for scattered $\alpha $ -particles.
An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze.$ Then the distance of closest approach for the alpha nucleus will be proportional to
An alpha nucleus of energy $\frac{1}{2}mv^2$ bombards a heavy nuclear target of charge $Ze.$ Then the distance of closest approach for the alpha nucleus will be proportional to
- [AIEEE 2006]
Assertion: The specific charge of positive rays is not constant.
Reason: The mass of ions varies with speed.
Assertion: The specific charge of positive rays is not constant.
Reason: The mass of ions varies with speed.
- [AIIMS 1999]