
If the atom \[_{100}F{m^{257}}\] follows the Bohr model and the radius of \[_{100}F{m^{257}}\] is $n$ times the Bohr radius, then find $n$ .
A. \[100\]
B. \[200\]
C. \[4\]
D. $\dfrac{1}{4}$
Answer
163.8k+ views
Hint:In the Bohr atomic model, electrons circle the nucleus in well-defined circular orbits. The quantum number $n$ , an integer, is used to identify the orbits. To find the required value of $n$ we will write the radius of \[_{100}F{m^{257}}\] in terms of Bohr radius and equate with the given value of radius of \[_{100}F{m^{257}}\] .
Formula Used:
Radius of an atom,
$r = \dfrac{{{m^2}{r_o}}}{Z}$
Complete step by step solution:
Given: $r = n{r_o}$...........(1)
Where $r$ is the radius of an atom (here, that of $Fm$ ) and ${r_o}$ is the Bohr radius. The synthetic element fermium has an atomic number of \[100\] and has the symbol $Fm$. Although pure fermium metal has not yet been created, it is an actinide and the heaviest element that can be created by neutron bombardment of lighter elements. As a result, it is the last element that can be synthesised in macroscopic amounts. We know that, radius of an atom can be written as,
$r = \dfrac{{{m^2}{r_o}}}{Z}$...........(2)
Now, electronic configuration of \[_{100}F{m^{257}}\] is \[2,{\text{ }}8,{\text{ }}18,{\text{ }}32,{\text{ }}50\] , that is, there are five number of orbits in \[_{100}F{m^{257}}\] . This implies that $m = 5$.
Putting the known values in equation (2), we get,
$r = \dfrac{{{5^2}{r_o}}}{{100}}$
Solving this we get,
$r = \dfrac{1}{4}{r_o}$
Comparing equation (1) and equation (2), we get,
$\therefore n = \dfrac{1}{4}$
Hence, option D is the answer.
Note: The Bohr model, often known as a planetary model, states that the electrons orbit the atom's nucleus in fixed, permitted paths. The energy of the electrons is fixed when it is in one of these orbits.
Formula Used:
Radius of an atom,
$r = \dfrac{{{m^2}{r_o}}}{Z}$
Complete step by step solution:
Given: $r = n{r_o}$...........(1)
Where $r$ is the radius of an atom (here, that of $Fm$ ) and ${r_o}$ is the Bohr radius. The synthetic element fermium has an atomic number of \[100\] and has the symbol $Fm$. Although pure fermium metal has not yet been created, it is an actinide and the heaviest element that can be created by neutron bombardment of lighter elements. As a result, it is the last element that can be synthesised in macroscopic amounts. We know that, radius of an atom can be written as,
$r = \dfrac{{{m^2}{r_o}}}{Z}$...........(2)
Now, electronic configuration of \[_{100}F{m^{257}}\] is \[2,{\text{ }}8,{\text{ }}18,{\text{ }}32,{\text{ }}50\] , that is, there are five number of orbits in \[_{100}F{m^{257}}\] . This implies that $m = 5$.
Putting the known values in equation (2), we get,
$r = \dfrac{{{5^2}{r_o}}}{{100}}$
Solving this we get,
$r = \dfrac{1}{4}{r_o}$
Comparing equation (1) and equation (2), we get,
$\therefore n = \dfrac{1}{4}$
Hence, option D is the answer.
Note: The Bohr model, often known as a planetary model, states that the electrons orbit the atom's nucleus in fixed, permitted paths. The energy of the electrons is fixed when it is in one of these orbits.
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