Consider $3^{r d}$ orbit of $H e^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, $Z = 2$ and h (Planck's Constant) $= 6.6 \times 10^{- 34} J s]$

0.73 \times 10^{6} m / s

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3.0 \times 10^{8} m / s

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2.92 \times 10^{6} m / s

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1.46 \times 10^{6} m / s

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Solution

The correct option is D $1.46 \times 10^{6} m / s$
$v = (2.19 \times 10^{6}) \frac{Z}{n}$

$v = (2.19 \times 10^{6}) \frac{2}{3}$

$v = 1.46 \times 10^{6} m / s$

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Consider $3^{r d}$ orbit of $H e^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, Z = 2 and h(Planck’s Constant) = $6.6 \times 10^{- 34}$ J s]

What is the speed of electron in the second Bohr's orbit of Hydrogen atom? $(T a k e v_{0} = 2.2 \times 10^{6} m / s)$ .

The circumference of the first Bohr orbit in H atom is $3.322 \times 10^{- 10} m$ . What is the velocity of the electron of this orbit?

Q. Work function of a metal is 3eV. An electromagnetic radiation of 200 nm is incident on metal surface, calculate the maximum speed of photoelectrons ejected. (Take the value of h=

6.6 \times 10^{- 34} J s)

Q. If the velocity of the electron in first orbit of hydrogen atom is

2.18 \times 10^{6} m/s

, what is its value in third orbit?

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Consider 3rd orbit of He+ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given K=9×109 constant, Z=2 and h (Planck's Constant) =6.6×10−34Js]

The correct option is D 1.46×106m/sv=(2.19×106)Znv=(2.19×106)23v=1.46×106m/s

Consider $3^{r d}$ orbit of $H e^{+}$ (Helium), using non-relativistic approach, the speed of electron in this orbit will be [given $K = 9 \times 10^{9}$ constant, $Z = 2$ and h (Planck's Constant) $= 6.6 \times 10^{- 34} J s]$

The correct option is D $1.46 \times 10^{6} m / s$
$v = (2.19 \times 10^{6}) \frac{Z}{n}$

$v = (2.19 \times 10^{6}) \frac{2}{3}$

$v = 1.46 \times 10^{6} m / s$