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A triply ionized beryllium ( Be ³⁺ ) has the same orbital radius as the ground state of hydrogen. Then the quantum state n of Be ³⁺ is

Options

  1. An=1
  2. Bn=2
  3. Cn=3
  4. Dn=4

Correct answer

B. n=2

Step-by-step solution

Radius of n^ th orbit in hydrogen like atoms is r_n= a₀ n^2 Z where a₀ is the Bohr's radius For hydrogen atom, Z=1 r₁=a₀ ( n=1 for ground state ) For Be ³⁺, Z=4 r_n= a₀ n^2 4 According to given problem r₁=r_na₀= n^2 a₀ 4 n=2

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