## Related questions with answers

A hydrogen atom is in its first excited state (n = 2). Calculate the kinetic energy of the electron.

Solutions

VerifiedTo calculate the kinetic energy of electron $K$, we can first find it's speed to determine if we have to use relativistic or non-relativistic relations. From (4) we have

$\begin{align*} v_2 &= \frac{p_2}{m_e} \\ v_2 &= \frac{ 9.97 \cdot 10^{-25} \: \mathrm{ kg \cdot m/s } }{ 9.11 \cdot 10^{-31} \:\mathrm{kg} } \\ v_2 &= 1.09 \cdot 10^6 \: \mathrm{m/s}. \tag{6} \end{align*}$

Although speed of electron is very high (for our standards), it is very small compared to speed of light. Therefore, all our non-relativistic calculations are justified. Kinetic energy of electron (in state $n=2$) is

$\begin{align*} K &= \frac{1}{2} m_e v_2^2 \\ K &= \frac{1}{2} 9.11 \cdot 10^{-31} \:\mathrm{kg} \cdot \left( 1.09 \cdot 10^6 \: \mathrm{m/s} \right)^2 \\ K &= 5.45 \cdot 10^{-19} \: \mathrm{ J } \cdot \frac{1 \: \mathrm{eV}}{ 1.6 \cdot 10^{-19} \: \mathrm{J} } \tag{7} \\ K &= \boxed{ 3.40 \: \mathrm{eV} }. \tag{8} \end{align*}$

State of the hydrogen atom: $n=2$,

We need to calculate the kinetic energy of the electron $T$.

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