MHT CET 2025 23rd April Morning Shift | Dual Nature of Radiation Question 9 | Physics

The energy that should be added to an electron to reduce its de-Broglie wavelength from $\lambda$ to $\frac{\lambda}{2}$ is $n$ times the initial energy. The value of ' $n$ ' is

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

When the electron orbiting in hydrogen atom goes from one orbit to another orbit (principal quantum number $=n$ ), the de-Broglie wavelength ( $\lambda$ ) associated with it is related to $n$ as

$\lambda \propto \frac{1}{\mathrm{n}^2}$

$\lambda \propto n^2$

$\lambda \propto \frac{1}{n}$

$\lambda \propto \mathrm{n}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

Photoelectric emission takes place from a certain metal at threshold frequency $v$. If the radiation of frequency $4 v$ is incident on the metal plate, the maximum velocity of the emitted photoelectrons will be ( $m=$ mass of photoelectron, $h=$ Planck's constant)

$\sqrt{\frac{6 \mathrm{hv}}{\mathrm{m}}}$

$\sqrt{\frac{3 h v}{m}}$

$\sqrt{\frac{h v}{m}}$

$\sqrt{\frac{5 \mathrm{hv}}{\mathrm{m}}}$

MHT CET 2025 22nd April Evening Shift

MCQ (Single Correct Answer)

-0

The de-Broglie wavelength of a neutron at $27^{\circ} \mathrm{C}$ is ' $\lambda_0$ '. What will be its wavelength at $927^{\circ} \mathrm{C}$ ?

$\frac{\lambda_0}{4}$

$\frac{\lambda_0}{3}$

$\frac{\lambda_0}{2}$

$\frac{3 \lambda_0}{2}$

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