For a hydrogen-like atom, classify these electron transitions by whether they result in the absorption or emission of light. Ignoring sign, which transition is associated with the greatest energy change? n = 3 to n = 5 n = 3 to n = 2 n = 2 to n = 1 n = 1 to n = 3

## Answer

**1)**

**when there is transition from higher state to lower state, it involves emission**

**when there is transition from lower state to higher state, it involves absorption**

**So, Absorption are:**

**n=3 to n=5**

**n=1 to n=3**

**emission are:**

**n=2 to n=1**

**n=3 to n=2**

**2)**

**a)**

**Here photon will be captured and it will excite the atom**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**R is Rydberg constant. R = 1.097*10^7**

**1/lambda = – R* (1/nf^2 – 1/ni^2)**

**1/lambda = – 1.097*10^7* (1/5^2 – 1/3^2)**

**lambda = 1.29*10^-6 m**

**lambda = 1290 nm**

**use:**

**E = h*c/lambda**

**=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.29*10^-6 m)**

**= 1.541*10^-19 J**

**b)**

**Here photon will be emitted**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**R is Rydberg constant. R = 1.097*10^7**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**1/lambda = 1.097*10^7* (1/2^2 – 1/3^2)**

**lambda = 6.563*10^-7 m**

**lambda = 656 nm**

**use:**

**E = h*c/lambda**

**=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(6.563*10^-7 m)**

**= 3.029*10^-19 J**

**Since this is emitted energy, sign will be negative**

**So,**

**E = 3.029*10^-19 J**

**C)**

**Here photon will be emitted**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**R is Rydberg constant. R = 1.097*10^7**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**1/lambda = 1.097*10^7* (1/1^2 – 1/2^2)**

**lambda = 1.215*10^-7 m**

**lambda = 122 nm**

**use:**

**E = h*c/lambda**

**=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.215*10^-7 m)**

**= 1.635*10^-18 J**

**Since this is emitted energy, sign will be negative**

**So,**

**E = 1.635*10^-18 J**

**D)**

**Here photon will be captured and it will excite the atom**

**1/lambda = R* (1/nf^2 – 1/ni^2)**

**R is Rydberg constant. R = 1.097*10^7**

**1/lambda = – R* (1/nf^2 – 1/ni^2)**

**1/lambda = – 1.097*10^7* (1/3^2 – 1/1^2)**

**lambda = 1.032*10^-7 m**

**lambda = 103 nm**

**use:**

**E = h*c/lambda**

**=(6.626*10^-34 J.s)*(3.0*10^8 m/s)/(1.032*10^-7 m)**

**= 1.926*10^-18 J**

**Clearly energy is largest for d**

**Answer: n=1 to n=3**