Question

name: _____________ pd: ____ chemistry u3l14_q2 ct module day 3 iiw
the photoelectron spectrum for an unknown element is shown above.
** (a) based on the photoelectron spectrum, identify the unknown element and write its electron configuration.
** (b) consider the element in the periodic table that is directly to the right of the element identified in part (a). would the 1s peak of this element appear to the left of, the right of, or in the same position as the 1s peak of the element in part (a)? explain your reasoning.

Explanation:

Step1: Count peak groups (energy levels)

The spectrum has 4 peak groups, meaning 4 energy levels (n=1 to n=4).

Step2: Determine electron counts per level

From left (highest binding energy) to right:

• n=1 (1s): 2 electrons (1 peak, typical for 1s)
• n=2 (2s, 2p): 8 electrons (2 peaks, 2+6)
• n=3 (3s, 3p): 8 electrons (2 peaks, 2+6)
• n=4 (4s): 1 electron (1 peak)

Total electrons: 2+8+8+1=19.

Step3: Identify element and electron config

19 electrons = atomic number 19 → Potassium (K). Electron configuration: $1s^22s^22p^63s^23p^64s^1$.

Step4: Analyze right neighbor (Ca, Z=20)

Ca has more protons than K, so higher nuclear charge. 1s electrons are closer to nucleus, so higher binding energy (left shift in spectrum).

Answer:

(a) Element: Potassium (K); Electron configuration: $1s^22s^22p^63s^23p^64s^1$
(b) The 1s peak of Ca appears to the left of K's 1s peak because Ca has a higher nuclear charge, increasing the binding energy of 1s electrons.