Question

An electron is in excited state in a hydrogen like atom. It has a total energy of $-3.4eV$. Find the kinetic energy of the electron is E and its De Broglie wavelength is $\lambda$.

• A. $E=6.8eV,\lambda =6.6\times {10}^{-10}m$
• B. $E=3.4eV,\lambda =6.6\times {10}^{-10}m$
• C. $E=3.4eV,\lambda =6.6\times {10}^{-11}m$
• D. $E=6.8eV,\lambda =6.6\times {10}^{-11}m$

Answer 1

Answer: (B)

$E=3.4eV,\lambda =6.6\times {10}^{-10}m$

As we know that

$TE=-KE$

And, ATQ, Energy $=-3.4eV$

Therefore, KInetic energy $3.4eV$

Now, energy of an orbit n is given by,

$E_n=\frac{-13.4}{n^2}$

ATQ, $E_n=TE=-3.4eV-3.4=\frac{-13.4}{n^2}n=2$
Velocity of an electron in 2nd orbit of hydrogen is given by,

$V=\frac{2\pi k\left(i\right)e^2}{\left(2\right)h}=1.09\times {10}^{6}m/s$

So, de Broglie wavelength is given by,

$\lambda =\frac{h}{mv}=\frac{6.63\times {10}^{-34}}{9.11\times 1.09\times {10}^6\times {10}^{-31}}=6.67\times {10}^{-10}m$