Question

Using Bohr atomic model, drive expression for calculating the radius of orbits (5) in He+. Using this expression, calculate the radius of fourth orbit of He+ ion.

Answer 1

radius is 25/2 pm of 5th orbit
radius is 8 pm of 4th orbit

Answer 2

Answer : The radius of fourth orbit of hydrogen like atom $(He^+)$ is, $423.2pm$

Solution :

Derivation for calaculating the radius of orbits of hydrogen atom and hydrogen like atom, $He^+$.

According to the Bohr's theory of hydrogen like atom, the electrons move around the nucleus in a stationary circular orbit.

Two forces are applied which are attractive forces and centripetal forces.

Attractive force : $F_A=\frac{Ze^2}{4\pi \epsilon_or^2}$   .......(1)

Centripetal force : $F_c=\frac{mv^2}{r}$     ........(2)

Now for electrons to remain stationary at a position,

$\frac{mv^2}{r}=\frac{Ze^2}{4\pi \epsilon_or^2}$

$mv^2=\frac{Ze^2}{4\pi \epsilon_or}$      .........(3)

Now from the angular momentum condition,

$mvr=\frac{nh}{2\pi}$

or,

$mv^2=\frac{n^2h^2}{4\pi^2r^2m}$     ..........(4)

Now equating equation 4 in 3, we get

$r_n=\frac{n^2\times 0.529A^o}{Z}$

$r_n=\frac{n^2\times 52.9}{Z}$   (in pm)

Formula used for the radius of the $n^{th}$ orbit will be,

$r_n=\frac{n^2\times 52.9}{Z}$   (in pm)

where,

$r_n$ = radius of $n^{th}$ orbit

n = number of orbit  = 4

Z = atomic number  of helium = 2

Now we have to calculate the radius of fourth orbit of hydrogen like atom $(He^+)$

Radius of first orbit of $(He^+)$ :

$r_4=\frac{(4)^2\times 52.9}{2}=423.2pm$

Therefore, the radius of fourth orbit of hydrogen like atom $(He^+)$ is, $423.2pm$