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Q 1. The kinetic energy of a particle is equal to the energy of a photon. The particle moves at {tex} 5 \% {/tex} of the speed of light. The ratio of the photon wavelength to the de Broglie wavelength of the particle is [No need to use relative formula for the particle {tex} ] {/tex}

40

4

2

80

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Q 2. Work function of nickel is {tex}5.01 eV{/tex}. When ultraviolet radiation of wavelength {tex} 200 \mathrm { Å } {/tex} is incident on it, electrons are emitted. What will be the maximum velocity of emitted electrons?

{tex} 3 \times 10 ^ { 8 } \mathrm { ms } ^ { - 1 } {/tex}

{tex} 6.46 \times 10 ^ { 5 } \mathrm { ms } ^ { - 1 } {/tex}

{tex} 10.36 \times 10 ^ { 5 } \mathrm { ms } ^ { - 1 } {/tex}

{tex} 8.54 \times 10 ^ { 6 } \mathrm { ms } ^ { - 1 } {/tex}

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Q 3. The KE of the photoelectrons is {tex} E {/tex} when the incident wavelength is {tex} \lambda / 2 . {/tex} The KE becomes {tex} 2 E {/tex} when the incident wavelength is {tex} \lambda / 3 . {/tex} The work function of the metal is

{tex} h c / \lambda {/tex}

{tex} 2 h c / \lambda {/tex}

{tex} 3 h c / \lambda {/tex}

{tex} h c / 3 \lambda {/tex}

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Q 4. The energy of a photon is equal to the kinetic energy of a proton. The energy of photon is {tex} E . {/tex} Let {tex} \lambda _ { 1 } {/tex} be the de Broglie wavelength of the proton and {tex} \lambda _ { 2 } {/tex} be the wavelength of the photon. Then, {tex} \lambda _ { 1 } / \lambda _ { 2 } {/tex} is proportional to

{tex} E ^ { 0 } {/tex}

{tex} E ^ { 1 / 2 } {/tex}

{tex} E ^ { - 1 } {/tex}

{tex} E ^ { - 2 } {/tex}

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Q 5. Find the ratio of de Broglie wavelength of a proton and an {tex} \alpha {/tex} -particle which have been accelerated through same potential difference

{tex} 2 \sqrt { 2 }: 1 {/tex}

{tex}3:2{/tex}

{tex} 3 \sqrt { 2 }: 1 {/tex}

{tex}2:1{/tex}

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