Question

given that in the hydrogen atom the transition energy for n= 1 and n =2 rydberg states is 10.2 electron volt.the energy for the same transition in be3+ is

Answer 1

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Answer 2

Answer: The energy required to excite an electron from n = 1 to n = 2 is 163.2eV.

Explanation:

The formula used to calculate the energy of excitation is given as follows:

$E_n=\frac{-13.6\times Z^2}{n^2}eV$

where,

Z = atomic number = 4 (for beryllium)

n = Energy level

The transition of ionization energy is from $1\rightarrow 2$

Putting values in above equation, we get:

$E_{1\rightarrow 2}=E_2-E_1$

$E_{1\rightarrow 2}=(\frac{-13.6\times 4^2}{2^2})-(\frac{-13.6\times 4^2}{1^2})\\\\E_{1\rightarrow 2}=(-54.4+217.6)eV=163.2eV$

Hence, the energy required to excite an electron from n = 1 to n = 2 is 163.2eV.