Question

According to Bohr's theory, the energy in the $n^{th}$ orbit is $E_n=\frac{-2.176\times {10}^{-18}\times Z^2}{n^2}J$. The longest wavelength of light that will be needed to remove an electron from the 3rd orbit of $H{e}^{+}$ ion is:

• A. $\lambda =2055A^o$
• B. $\lambda =2154A^o$
• C. $\lambda =2164A^o$
• D. $\lambda =2004A^o$

Answer 1

The correct option is A $\lambda =2055A^o$

According to Bohr's model, the energy of an electron in the $n^{th}$ orbit of $H{e}^{+}$ ion is $E_n=\frac{-2.176\times {10}^{-18}\times Z^2}{n^2}J$
For $H{e}^{+}$, $Z=2$.
$E_3=\frac{-2.176\times {10}^{-18}\times 2^2}{3^2}J$$=-9.67\times {10}^{-19}Joules$
$\lambda =\frac{hc}{E}=\left(\frac{6.626\times {10}^{-34}\times 3\times {10}^8}{9.67\times {10}^{-19}}\right)m=2.055\times {10}^{-7}m=2055A^o$
Hence, the longest wavelength of light that will be needed to remove an electron from the 3rd orbit of $H{e}^{+}$ ion is $2055A^o$.