Rubidium has an atomic number of 37 and a mass number of 85. How many protons, neutrons, and electrons does an ion of rubidium with a 1+ charge have? (1 point)

Question

37 protons, 48 neutrons, and 38 electrons

48 protons, 37 neutrons, and 36 electrons

37 protons, 48 neutrons, and 36 electrons

48 protons, 37 neutrons, and 38 electrons

Answer 1

To determine the number of protons, neutrons, and electrons in a rubidium ion with a 1+ charge, we can follow these steps:

1. Protons: The atomic number of rubidium (Rb) is 37, which means it has 37 protons.

2. Neutrons: The mass number of rubidium is 85. The number of neutrons can be calculated as: \[ \text{Neutrons} = \text{Mass number} - \text{Atomic number} = 85 - 37 = 48. \]

3. Electrons: A neutral atom of rubidium has the same number of electrons as protons. Thus, it has 37 electrons. However, since the rubidium ion has a 1+ charge, it has lost one electron, resulting in: \[ \text{Electrons} = 37 - 1 = 36. \]

Putting it all together, a rubidium ion with a 1+ charge has:

• 37 protons,
• 48 neutrons, and
• 36 electrons.

So, the correct answer is: 37 protons, 48 neutrons, and 36 electrons.