Question

A hydrogen-1 atom in the ground state absorbs a UV photon. As a result, the 1H atom is ionized – it breaks into a proton moving with a speed of 1.15×103 m/s and an electron moving with a speed of 2.11×106 m/s in the opposite direction. Calculate the wavelength of the UV photon. [Note: masses of electron and proton are available in the textbook and on the internet; use masses in kg.]

Answer 1

Answer:

$\lambda = 390 nm$

Explanation:

Here we can use momentum conservation to find the initial wavelength of UV photon

so the equation for momentum conservation is given as

$\frac{h}{\lambda} = m_1v_1 + m_2v_2$

we will have

$v_1 = 1.15 \times 10^3 m/s$

$m_1 = 1.67 \times 10^{-27} kg$

$v_2 = -2.11 \times 10^6 m/s$

$m_2  = 9.11 \times 10^{-31} kg$

so we will have

$\frac{h}{\lambda} = (1.67 \times 10^{-27})(1.15 \times 10^3) - (9.11 \times 10^{-31})(2.11 \times 10^6)$

$\frac{h}{\lambda} = -1.71 \times 10^{-27}$

$\lambda = 390 nm$