Energy liberated in the de-excitation of hydrogen atom from 3 rd level to 1 st level falls on a photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen like

Question

Energy liberated in the de-excitation of hydrogen atom from $3^{rd}$ level to $1^{st}$ level falls on a photo-cathode. Later when the same photo-cathode is exposed to a spectrum of some unknown hydrogen like gas, excited to $2^{nd}$ energy level, it is found that the deBroglie wavelength of the fastest photoelectrons, now ejected has become one third. For this new gas, difference of energies of $2^{nd}$ Lyman line and $1^{st}$ Balmer line is found to be 3 times the ionization potential of the hydrogen atom. Select the correct statement(s):

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Answer:

BC

Solution:

$E_{0} Z^{2} (1 - \frac{1}{9}) - E_{0} Z^{2} (\frac{1}{4} - \frac{1}{9}) Z \frac{λ _{1}}{λ _{2}} K E_{1} K E_{2} K E = 3 E_{0} = 2 = 3 = E_{0} (1 - \frac{1}{9}) - ϕ = E_{0} Z^{2} (1 - \frac{1}{4}) - ϕ \propto \frac{1}{λ ^{2}} = 8.5 e V$

TestHub · Accepted Answer

E 0 ​ Z 2 ( 1 − 9 1 ​ ) − E 0 ​ Z 2 ( 4 1 ​ − 9 1 ​ ) Z λ 2 ​ λ 1 ​ ​ K E 1 ​ K E 2 ​ K E ​ = 3 E 0 ​ = 2 = 3 = E 0 ​ ( 1 − 9 1 ​ ) − ϕ = E 0 ​ Z 2 ( 1 − 4 1 ​ ) − ϕ ∝ λ 2 1 ​ = 8.5 e V ​

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