Cách giải bình thường:
Phải mò trạng thái dừng. Nhưng đưa ra nhận xét
Ở mức M (n = 3) có 3 vạch: (3 -> 2); (3 -> 1); (2 -> 1).
Vậy mức thỏa mãn 6 vạch phải lớn hơn n = 3. Thử với mức n = 4 (N) khi đó có các vạch:
(4 -> 3); (4 -> 2); (4 -> 1); (3 -> 2); (3 -> 1); (2 -> 1) tất cả là 6 vạch => chọn N.
Cách giải nhanh:
Nhận xét: 6 = 1+2+3 => trạng thái dừng cao nhất mà nguyên tử chỉ phát ra được 6 vạch là 3+1 = 4. Mức N.
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