Answer:
(1) The wavelength of the proton is 1.366 x 10⁻¹⁵ m
(2) 2p( l = 1, ml = -1,0,+1)
3s( n = 3, l = 0, ml = 0)
5d ( l = 2, ml = -2,-1,0,+1,+2)
Step-by-step explanation:
Given;
mass of the proton; m = 1.673 x 10⁻²⁴ g = 1.673 x 10⁻²⁷ kg
velocity of the proton, v = 2.9 x 10⁸ m/s
The wavelength of the proton is calculated by applying De Broglie's equation;
![\lambda = (h)/(mv)](https://img.qammunity.org/2021/formulas/chemistry/high-school/rtgljpzluy996qdr4zfecdwujfugek9oci.png)
where;
h is Planck's constant = 6.626 x 10⁻³⁴ J/s
Substitute the given values and solve for wavelength of the proton;
![\lambda = (h)/(mv)\\\\ \lambda = ((6.626*10^(-34)))/((1.673*10^(-27))(2.9*10^8))\\\\\lambda = 1.366 *10^(-15) \ m](https://img.qammunity.org/2021/formulas/physics/college/tjms3046x2slihwebu0qzou60typf2a5ao.png)
(2) the values for the quantum numbers associated with the following orbitals is given by;
n, which represents Principal Quantum number
which represents Azimuthal Quantum number
which represents Magnetic Quantum number
(a) 2p (number of orbital = 3):
![l= 1\\m_l = -1,0,+1](https://img.qammunity.org/2021/formulas/physics/college/532obui0yev5h0ycva62pddksej3csfjme.png)
(b) 3s (number of orbital = 1):
![n= 3\\l=0\\m_l= 0](https://img.qammunity.org/2021/formulas/physics/college/sljsutayw9m0t8q378c60v95a5xexw9cxd.png)
(c) 5d (number of orbital = 5)
![l=2\\m_l = 2, -1, 0, +1, +2](https://img.qammunity.org/2021/formulas/physics/college/xl6kazbzy5rrhyi82reox3ri23x9wu9au5.png)