Question # 1

The de-Broglie wavelength of the particle of mass m and energy E is:

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ℎ

𝜆 = ℎ√2𝑚𝐸

𝜆 = 1

Of electron of 50 keV strike a heavy target. Then radiation emitted by target will be

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Visible light

Radio waves

Ultraviolet

None of these

The Balmer series is found in the spectrum of:

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Hydrogen

Nitrogen

Oxygen

All

The minimum energy required to remove an electron is called:

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Stopping potential

Work function

Kinetic energy

None of these

The ratio of the longest and shortest wavelength of the Lyman series is approximately:

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4/3

9/4

9/5

16/7

In photoelectric effect experiment, stopping potential depend upon

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Intensity of light

Frequency of light

Photoelectric current

Both A and B

In which region of the electromagnetic spectrum does the Lyman series of hydrogen atom lie?

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Infrared

Visible

Ultraviolet

X-rays

An electron and a proton are accelerated through the same potential. If their masses are me and mp respectively, then the ratio of their de-Broglie wavelength is:

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1

mp/me

me/mp

Ultraviolet radiation of 6.2 eV falls on an aluminium surface having work function ∅ = 𝟒. 𝟐 𝒆𝑽.

The kinetic energy of the fastest electron emitted is:

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4 eV

2 eV

2.2 eV

1.2 eV

Light elements do not emit X-rays because

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Electrons in it have high binding energy

These materials are non- material

There is a small difference in their energy shells

Electrons in it require very large energy to remove from these materials

When an electron in an atom goes from a lower to higher its:

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K.E. increases, P.E. decreases

K.E. increases

P.E increases

K.E. decrease, P.E. increases

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MDCAT Physics Chapter 12 Atomic spectra Induction Online Test MCQs With Answers

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